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Unformatted text preview: Lecture 26 We continue our study of forms with compact support. Let us begin with a review. Let U ∈ R n be open, and let ω = f I ( x 1 , . . . , x n ) dx I , (5.48) I where I = ( i 1 , . . . , i k ) is strictly increasing and dx I = dx i 1 ∧ ··· ∧ dx i k . Then ω is compactly supported ⇐⇒ every f I is compactly supported . (5.49) By definition, supp f I = { x ∈ U : f I ( x ) = 0 } . (5.50) We assume that the f I ’s are C 2 maps. Notation. Ω c k ( U ) = space of compactly supported differentiable kforms on U . (5.51) Now, let ω ∈ Ω n c ( U ) defined by ω = f ( x 1 , . . . , x n ) dx 1 ∧ ··· ∧ dx n , (5.52) where f ∈ Ω 0 c ( U ). Then ω = f ( x 1 , . . . , x n ) dx 1 ∧ ··· ∧ dx n . (5.53) R n R n Last time we proved the Poincare Lemma for open rectangles R in R n . We assumed that ω ∈ Ω c n (Int R ). That is, we assumed that ω ∈ Ω c n ( R n ) such that supp ω ⊂ Int R . We showed that for such ω the following two conditions are equivalent: 1. R n ω = 0, 2. There exists a µ ∈ Ω n − 1 (Int R ) such that dµ = 0. c Definition 5.9. Whenever ω ∈ Ω k ( U ) and ω = dµ for some µ ∈ Ω k − 1 ( U ), we say that ω is exact . Definition 5.10. Whenever ω ∈ Ω k ( U ) such that dω = 0, we say that ω is closed . Observe that ω ∈ Ω n c ( U ) = dω = 0 . (5.54) ⇒ Now we prove the Poincare Lemma for open connected subsets of R n ....
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 Fall '04
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 Topology, Poincare Lemma

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