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Unformatted text preview: Lecture 35 Before moving on to integration, we make a few more remarks about orientations. Let X, Y be oriented manifolds. A diffeomorphism f : X Y is orientation preserving if for every p X , the map df p : T p X T q Y (6.85) is orientation preserving, where q = f ( p ). Let V be open in X , let U be open in R n , and let : U V be a parameterization. Definition 6.32. The map is an oriented parameterization if it is orientation pre serving. Suppose is orientation reversing. Let A : R n R n be the linear map defined by A ( x 1 , . . . , x n ) = ( x 1 , x 2 , . . . , x n ) . (6.86) The map A is orientation reversing. Let U = A 1 ( U ), and define = A : U V . Both and A are orientation reversing, so is orientation preserving. Thus, for every point p X , there exists an oriented parameterization of X at p . 6.7 Integration on Manifolds Our goal for today is to take any n c ( X ) and define . (6.87) X First, we consider a special case: Let : U V be an oriented parameterization. Let U be open in R n , and let V be open in X . Take any c n ( V ). Then = , (6.88) V U where = n , where f C ( U ) and f ( x ) dx 1 dx = f. (6.89) U U Claim. The above definition for does not depend on the choice of oriented pa rameterization . 1 Proof. Let i : U i V, i = 1 , 2, be oriented parameterizations. Let n c ( V 1 V 2 ). Define U 1 , 2 = 1 ( V 1 V 2 ) , (6.90) 1 U 2 , 1 = 1 ( V 1 V 2 ) , (6.91) 2 which are open sets in R n ....
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 Fall '04
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