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Unformatted text preview: H/wk 9 (SOLUTIONS). Due Wednesday, April 4. 1. For n ≥ r ≥ 1 define the following function I : Ω r ( R n ) → Ω r 1 ( R n ) from the space of smooth rforms on R n to the space of smooth ( r 1)forms of R n . For ω = X i 1 <i 2 < ··· <i r ω i 1 ...i r dx i 1 ∧ ··· ∧ dx i r I ( ω ) = X i 1 <i 2 < ··· <i r r X j =1 ( Z 1 ( 1) j +1 t r 1 ω i 1 ...i r ( tx ) dt ) x i j dx i 1 ∧···∧ d dx i j ∧···∧ dx i r . (a) Prove that ω = I ( dω ) + d ( Iω ) for every rform ω on R n . (b) Conclude that every closed rform ω on R n (where r ≥ 1) is exact. That is, show that if dω = 0 then there is some ( r 1)form α such that ω = dα . Solution. (a) By linearity of all the formulas involved it suffices to prove that ω = I ( dω ) + d ( Iω ) for all ω of the form ω = f ( x ) dx i 1 ∧ ··· ∧ dx i r where i 1 < i 2 < ··· < i r . We have dω = X k 6 = i 1 ,...,i r ∂f ∂x k dx k ∧ dx i 1 ∧ ··· ∧ dx i r . Then I ( dω ) = X k 6 = i 1 ,...,i r Z 1 t r ∂f ∂x k ( tx ) dt x k dx i 1 ∧ ··· ∧ dx i r + X k 6 = i 1 ,...,i r r X j =1 ( 1) j +2 Z 1 t r ∂f ∂x k ( tx ) dt x i j dx k ∧ dx i 1 ∧ . . . d dx i j ∧ ··· ∧ dx i r . On the other hand, I ( ω ) = r X j =1 ( 1) j +1 Z 1 t r 1 f ( tx ) dt x i j dx i 1 ∧ . . ....
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This note was uploaded on 04/23/2010 for the course MATH Math 481 taught by Professor Kapovich during the Fall '08 term at University of Illinois at Urbana–Champaign.
 Fall '08
 kapovich
 Math

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