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lecture25

lecture25 - Lecture 25 5.1 The Poincare Lemma Let U be an...

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Lecture 25 5.1 The Poincare Lemma Let U be an open subset of R n , and let ω Ω k ( U ) be a k -form. 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 0 ( 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 = , 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 0 ( U ). Every µ Ω n 1 ( U ) can be written this way. c Applying d we obtain � � ∂f i = dx i ∧ · · · ∧ dx n . (5.22) ∂x j dx j dx 1 ∧ · · · ∧ i j 1

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lecture25 - Lecture 25 5.1 The Poincare Lemma Let U be an...

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