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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 k ( U ) be a k-form. We can write = a I dx I , I = ( i 1 , . . . , i k ), where each a I C ( U ). Note that k 0 ( U ) for each I. (5.17) c a I C We are interested in n c ( U ), which are of the form = fdx 1 dx n , (5.18) where f C ( U ). We define = 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 n c ( U ) . The following conditions are equivalent: 1. = 0 , U 2. = d , for some n 1 ( U ) . c In todays 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 n 1 ( U ). Specifically, define c = f i dx 1 (5.21) dx i dx n , i where each f i C ( U ). Every n 1 ( U ) can be written this way. c Applying d we obtain f i d = 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 d...
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- Fall '04