Differential Geometry 1
Homework 06
Let
α
∈
Ω
1
(
M
) be nowherezero and choose
ζ
∈
Vec
M
such that
α
(
ζ
) is
identically one.
1. Show that if
β
∈
Ω
k
+1
(
M
) and
α
∧
β
= 0 then
β
=
α
∧
γ
for some
γ
∈
Ω
k
(
M
). [
Hint
: Recall that
ζ
is an antiderivation.]
2. Assume that
α
∧
d
α
= 0 and choose
γ
∈
Ω
1
(
M
) such that d
α
=
α
∧
γ
.
Show that
γ
∧
d
γ
∈
Ω
3
(
M
) is closed; show further that if also d
α
=
α
∧
γ
then
γ
∧
d
γ

γ
∧
d
γ
is exact. [
Hint
: First show
α
∧
d
γ
= 0.]
Remark
: This shows that
α
gives rise to a welldefined cohomology class
[
γ
∧
d
γ
]
∈
H
3
dR
(
M
).
Solutions
(1) Very brief: If
α
∧
β
= 0 then 0 =
ζ
(
α
∧
β
) = (
ζ α
)
∧
β

α
∧
(
ζ β
)
as
is an antiderivation; as
ζ α
≡
1 it follows that
β
=
α
∧
(
ζ β
).
(2) From above, d
α
=
α
∧
γ
for some
γ
∈
Ω
1
(
M
).
Now 0 = d
2
α
=
d(
α
∧
γ
) = d
α
∧
γ

α
∧
d
γ
=

α
∧
d
γ
because d
α
∧
γ
= (
α
∧
γ
)
∧
γ
= 0;
from above again, it follows that d
γ
=
α
∧
ε
for some
ε
∈
Ω
1
(
M
).
To see that
γ
∧
d
γ
is closed, compute:
d(
γ
∧
d
γ
) = d
γ
∧
d
γ
= (
α
∧
ε
)
∧
(
α
∧
ε
) = 0
.
To see that
γ
∧
d
γ

γ
∧
d
γ
is exact, first let d
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
This is the end of the preview.
Sign up
to
access the rest of the document.
 Spring '09
 Robinson
 Topology, Algebraic Topology, Differential form, dΓ, Cohomology

Click to edit the document details