5Differential forms and de Rham cohomology

III Differential Geometry



5.1 Differential forms
We are now going to restrict our focus to a special kind of tensors, known as
differential forms. Recall that in
R
n
(as a vector space), an alternating
n
-linear
map tells us the signed volume of the parallelepiped spanned by
n
vectors.
In general, a differential
p
-form is an alternating
p
-linear map on the tangent
space at each point, so it tells us the volume of an “infinitesimal
p
-dimensional
parallelepiped”.
In fact, we will later see than on an (oriented)
p
-dimensional manifold, we
can integrate a
p
-form on the manifold to obtain the “volume” of the manifold.
Definition (Differential form). We write
p
(M) = C
(M, Λ
p
T
M) = {p-forms on M}.
An element of
p
(M) is known as a differential p-form.
In particular, we have
0
(M) = C
(M, R).
In local coordinates x
1
, · · · , x
n
on U we can write ω
p
(M) as
ω =
X
i
1
<...<i
p
ω
i
1
,...,i
p
dx
i
1
· · · dx
i
p
for some smooth functions ω
i
1
,...,i
p
.
We are usually lazy and just write
ω =
X
I
ω
I
dx
I
.
Example. A 0-form is a smooth function.
Example.
A 1-form is a section of
T
M
. If
ω
1
(
M
) and
X Vect
(
M
),
then ω(X) C
(M, R).
For example, if f is a smooth function on M , then df
1
(M) with
df(X) = X(f)
for all X Vect(M).
Locally, we can write
df =
n
X
i=1
a
i
dx
i
.
To work out what the a
i
’s are, we just hit this with the
x
j
. So we have
a
j
= df
x
j
=
f
x
j
.
So we have
df =
n
X
i=1
f
x
i
dx
i
.
This is essentially just the gradient of a function!
Example. If dim M = n, and ω
n
(M), then locally we can write
ω = g dx
1
· · · dx
n
.
for some smooth function
g
. This is an alternating form that assigns a real
number to n tangent vectors. So it measures volume!
If y
1
, · · · , y
n
is any other coordinates, then
dx
i
=
X
x
i
y
j
dy
j
.
So we have
ω = g det
x
i
y
j
i,j
dy
1
· · · dy
n
.
Now a motivating question is this given an
ω
1
(
M
), can we find some
f
0
(M) such that ω = df?
More concretely, let U R
2
be open, and let x, y be the coordinates. Let
ω = a dx + b dy
If we have w = df for some f , then we have
a =
f
x
, b =
f
y
.
So the symmetry of partial derivatives tells us that
a
y
=
b
x
. ()
So this equation () is a necessary condition to solve ω = df. Is it sufficient?
To begin with, we want to find a better way to express (
) without resorting
to local coordinates, and it turns out this construction will be very useful later
on.
Theorem (Exterior derivative). There exists a unique linear map
d = d
M,p
: Ω
p
(M)
p+1
(M)
such that
(i) On
0
(M) this is as previously defined, i.e.
df(X) = X(f) for all X Vect(M ).
(ii) We have
d d = 0 : Ω
p
(M)
p+2
(M).
(iii) It satisfies the Leibniz rule
d(ω σ) = dω σ + (1)
p
ω dσ.
It follows from these assumptions that
(iv)
d acts locally, i.e. if
ω, ω
0
p
(
M
) satisfy
ω|
U
=
ω
0
|
U
for some
U M
open, then dω|
U
= dω
0
|
U
.
(v) We have
d(ω|
U
) = (dω)|
U
for all U M.
What do the three rules tell us? The first rule tells us this is a generalization
of what we previously had. The second rule will turn out to be a fancy way of
saying partial derivatives commute. The final Leibniz rule tells us this d is some
sort of derivative.
Example. If we have
ω = a dx + b dy,
then we have
dω = da dx + a d(dx) + db dy + b d(dy)
= da dx + db dy
=
a
x
dx +
a
y
dy
dx +
b
x
dx +
b
y
dy
dy
=
b
x
a
y
dx dy.
So the condition () says dω = 0.
We now rephrase our motivating question if ω
1
(M) satisfies dω = 0,
can we find some
f
such that
ω
= d
f
for some
f
0
(
M
)? Now this has the
obvious generalization given any
p
-form
ω
, if d
ω
= 0, can we find some
σ
such that ω = dσ?
Example.
In
R
3
, we have coordinates
x, y, z
. We have seen that for
f
0
(
R
3
),
we have
df =
f
x
dx +
f
y
dy +
f
z
dz.
Now if
ω = P dx + Q dy + R dz
1
(R
3
),
then we have
d(P dx) = dP dx + P ddx
=
P
x
dx +
P
y
dy +
P
z
dz
dx
=
P
y
dx dy
P
z
dx dz.
So we have
dω =
Q
x
P
y
dx dy +
R
x
P
z
dx dz +
R
y
Q
z
dy dz.
This is just the curl! So d
2
= 0 just says that curl grad = 0.
Proof.
The above computations suggest that in local coordinates, the axioms
already tell use completely how d works. So we just work locally and see that
they match up globally.
Suppose
M
is covered by a single chart with coordinates
x
1
, · · · , x
n
. We
define d : Ω
0
(M)
1
(M) as required by (i). For p > 0, we define
d
X
i
1
<...<i
p
ω
i
1
,...,i
p
dx
i
1
· · · dx
i
p
=
X
dω
i
1
,...,i
p
dx
i
1
· · · dx
i
p
.
Then (i) is clear. For (iii), we suppose
ω = f dx
I
p
(M)
σ = g dx
J
q
(M).
We then have
d(ω σ) = d(fg dx
I
dx
J
)
= d(fg) dx
I
dx
J
= g df dx
I
dx
J
+ f dg dx
I
dx
J
= g df dx
I
dx
J
+ f(1)
p
dx
I
(dg dx
J
)
= (dω) σ + (1)
p
ω dσ.
So done. Finally, for (ii), if f
0
(M), then
d
2
f = d
X
i
f
x
i
dx
i
!
=
X
i,j
2
f
x
i
x
j
dx
j
dx
i
= 0,
since partial derivatives commute. Then for general forms, we have
d
2
ω = d
2
X
ω
I
dx
I
= d
X
dω
I
dx
I
= d
X
dω
I
dx
i
1
· · · dx
i
p
= 0
using Leibniz rule. So this works.
Certainly this has the extra properties. To claim uniqueness, if
: Ω
p
(
M
)
p+1
(M) satisfies the above properties, then
ω =
X
ω
I
dx
I
=
X
ω
I
dx
I
+ ω
I
dx
I
=
X
dω
I
dx
I
,
using the fact that = d on
0
(M) and induction.
Finally, if
M
is covered by charts, we can define d : Ω
p
(
M
)
p+1
(
M
) by
defining it to be the d above on any single chart. Then uniqueness implies this is
well-defined. This gives existence of d, but doesn’t immediately give uniqueness,
since we only proved local uniqueness.
So suppose
: Ω
p
(
M
)
p+1
(
M
) again satisfies the three properties. We
claim that
is local. We let
ω, ω
0
p
(
M
) be such that
ω|
U
=
ω
0
|
U
for some
U M
open. Let
x U
, and pick a bump function
χ C
(
M
) such that
χ 1 on some neighbourhood W of x, and supp(χ) U . Then we have
χ · (ω ω
0
) = 0.
We then apply to get
0 = (χ · (ω ω
0
)) = dχ (ω ω
0
) + χ(ω ω
0
).
But χ 1 on W . So dχ vanishes on W . So we must have
ω|
W
ω
0
|
W
= 0.
So ω = ω
0
on W .
Finally, to show that
=
d
, if
ω
p
(
M
), we take the same
χ
as before,
and then on x, we have
ω =
χ
X
ω
I
dx
I
= χ
X
ω
I
dx
I
+ χ
X
ω
I
dx
I
= χ
X
dω
I
dx
I
= dω.
So we get uniqueness. Since x was arbitrary, we have = d.
One useful example of a differential form is a symplectic form.
Definition
(Non-degenerate form)
.
A 2-form
ω
2
(
M
) is non-degenerate if
ω(X
p
, X
p
) = 0 implies X
p
= 0.
As in the case of an inner product, such an
ω
gives us an isomorphism
T
p
M T
p
M by
α(X
p
)(Y
p
) = ω(X
p
, Y
p
).
Definition
(Symplectic form)
.
A symplectic form is a non-degenerate 2-form
ω
such that dω = 0.
Why did we work with covectors rather than vectors when defining differential
forms? It happens that differential forms have nicer properties. If we have some
F C
(M, N) and g
0
(N) = C
(N, R), then we can form the pullback
F
g = g F
0
(M).
More generally, for x M , we have a map
DF |
x
: T
x
M T
F (x)
N.
This does not allow us to pushforward a vector field on
M
to a vector field of
N, as the map F might not be injective. However, we can use its dual
(DF |
x
)
: T
F (x)
N T
x
M
to pull forms back.
Definition
(Pullback of differential form)
.
Let
ω
p
(
N
) and
F C
(
M, N
).
We define the pullback of ω along F to be
F
ω|
x
= Λ
p
(DF |
x
)
(ω|
F (x)
).
In other words, for v
1
, · · · , v
p
T
x
M, we have
(F
ω|
x
)(v
1
, · · · , v
p
) = ω|
F (x)
(DF |
x
(v
1
), · · · , DF |
x
(v
p
)).
Lemma. Let F C
(M, N). Let F
be the associated pullback map. Then
(i) F
is a linear map
p
(N)
p
(M).
(ii) F
(ω σ) = F
ω F
σ.
(iii) If G C
(N, P ), then (G F )
= F
G
.
(iv) We have dF
= F
d.
Proof.
All but (iv) are clear. We first check that this holds for 0 forms. If
g
0
(N), then we have
(F
dg)|
x
(v) = dg|
F (x)
(DF |
x
(v))
= DF |
x
(v)(g)
= v(g F )
= d(g F )(v)
= d(F
g)(v).
So we are done.
Then the general result follows from (i) and (ii). Indeed, in local coordinates
y
1
, · · · , y
n
, if
ω =
X
ω
i
1
,...,i
p
dy
i
1
· · · dy
i
p
,
then we have
F
ω =
X
(F
ω
i
1
,...,i
p
)(F
dy
i
1
· · · dy
i
p
).
Then we have
dF
ω = F
dω =
X
(F
dω
i
1
,...,i
p
)(F
dy
i
1
· · · dy
i
p
).