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Unformatted text preview: Diﬀerential Geometry 1
Homework 06
Let α ∈ Ω1 (M ) be nowherezero and choose ζ ∈ VecM such that α(ζ ) is
identically one.
1. Show that if β ∈ Ωk+1 (M ) and α ∧ β = 0 then β = α ∧ γ for some
γ ∈ Ωk (M ). [Hint: Recall that ζ is an antiderivation.]
2. Assume that α ∧ dα = 0 and choose γ ∈ Ω1 (M ) such that dα = α ∧ γ .
Show that γ ∧ dγ ∈ Ω3 (M ) is closed; show further that if also dα = α ∧ γ
then γ ∧ dγ − γ ∧ dγ is exact. [Hint: First show α ∧ dγ = 0.]
Remark: This shows that α gives rise to a welldeﬁned cohomology class
3
[γ ∧ dγ ] ∈ HdR (M ).
Solutions
as (1) Very brief: If α ∧ β = 0 then 0 = ζ (α ∧ β ) = (ζ α) ∧ β − α ∧ (ζ β )
is an antiderivation; as ζ α ≡ 1 it follows that β = α ∧ (ζ β ). (2) From above, dα = α ∧ γ for some γ ∈ Ω1 (M ). Now 0 = d2 α =
d(α ∧ γ ) = dα ∧ γ − α ∧ dγ = −α ∧ dγ because dα ∧ γ = (α ∧ γ ) ∧ γ = 0;
from above again, it follows that dγ = α ∧ ε for some ε ∈ Ω1 (M ).
To see that γ ∧ dγ is closed, compute:
d(γ ∧ dγ ) = dγ ∧ dγ = (α ∧ ε) ∧ (α ∧ ε) = 0.
To see that γ ∧ dγ − γ ∧ dγ is exact, ﬁrst let dγ = α ∧ ε and note
γ ∧ α = γ ∧ α(= −dα), then compute:
γ ∧ dγ − γ ∧ dγ = (γ − γ ) ∧ dγ + γ ∧ d(γ − γ )
where the ﬁrst term on the right is given by
(γ − γ ) ∧ dγ = (γ − γ ) ∧ (α ∧ ε ) = (γ ∧ α − γ ∧ α) ∧ ε = 0
and where from
d((γ − γ ) ∧ γ ) = d(γ − γ ) ∧ γ − (γ − γ ) ∧ dγ = d(γ − γ ) ∧ γ − 0
1 we deduce (as d(γ − γ ) has even degree) that the second term is given by
γ ∧ d(γ − γ ) = d((γ − γ ) ∧ γ ) = d(γ ∧ γ );
thus
γ ∧ dγ − γ ∧ dγ = d(γ ∧ γ ).
Done! There are other possible lines of argument, but this one results in a
neat formula. Of course, once the formula is known it can be established a
little more directly by expanding d(γ ∧ γ ). 2 ...
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This note was uploaded on 07/08/2011 for the course MTG 6256 taught by Professor Robinson during the Spring '09 term at University of Florida.
 Spring '09
 Robinson

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