This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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 kform. 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...
View
Full
Document
 Fall '04
 unknown

Click to edit the document details