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Unformatted text preview: Lecture 25 5.1 The Poincare Lemma Let U be an open subset of R n , and let ω ∈ Ω k ( U ) be a kform. We can write ω = a I dx I , I = ( i 1 , . . . , i k ), where each a I ∈ C ∞ ( U ). Note that ω ∈ Ω k 0 ( U ) for each I. (5.17) c ⇐⇒ a I ∈ C ∞ We are interested in ω ∈ Ω n c ( U ), which are of the form ω = fdx 1 ∧ ··· ∧ dx n , (5.18) where f ∈ C ∞ ( U ). We define ω = f = fdx, (5.19) U U U the Riemann integral of f over U . Our goal over the next couple lectures is to prove the following fundamental the orem known as the Poincare Lemma. Poincare Lemma. Let U be a connected open subset of R n , and let ω ∈ Ω n c ( U ) . The following conditions are equivalent: 1. ω = 0 , U 2. ω = dµ , for some µ ∈ Ω n − 1 ( U ) . c In today’s lecture, we prove this for U = Int Q , where Q = [ a 1 , b 1 ] × ··· × [ a n , b n ] is a rectangle. Proof. First we show that (2) implies (1). Notation. (5.20) dx 1 ∧ ··· ∧ dx i ∧ ··· ∧ dx n ≡ dx 1 ∧ ··· ∧ dx i − 1 ∧ dx i +1 ∧ ··· ∧ dx n . Let µ ∈ Ω n − 1 ( U ). Specifically, define c µ = f i dx 1 ∧ ··· ∧ (5.21) dx i ∧ ··· ∧ dx n , i where each f i ∈ C ∞ ( U ). Every µ ∈ Ω n − 1 ( U ) can be written this way. c Applying d we obtain ∂f i dµ = dx i ∧ ··· ∧ dx n . (5.22) ∂x j dx j ∧ dx 1 ∧ ··· ∧ i j 1 Notice that if i = j , then the i, j th summand is zero, so ∂f i dµ...
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 Fall '04
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 Trigraph, dxn, dµ

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