HW06sol - Differential Geometry 1 Homework 06 Let α ∈...

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Unformatted text preview: Differential Geometry 1 Homework 06 Let α ∈ Ω1 (M ) be nowhere-zero 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 well-defined 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.

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HW06sol - Differential Geometry 1 Homework 06 Let α ∈...

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