lecture35

lecture35 - Lecture 35 Before moving on to integration we...

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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 φ : U → V be a parameterization. Definition 6.32. The map φ is an oriented parameterization if it is orientation pre- serving. Suppose φ 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 φ = φ A : U V . ◦ → Both φ and A are orientation reversing, so φ 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 ω ∈ Ω n c ( X ) and define ω. (6.87) X First, we consider a special case: Let φ : U V be an oriented parameterization. Let U be open in R n , and let V → be open in X . Take any ω ∈ Ω c n ( V ). Then ω = φ ∗ ω, (6.88) V U where φ ∗ ω = n , where f ∈ C ∞ ( U ) and f ( x ) dx 1 ∧ ··· ∧ dx φ ∗ ω = f. (6.89) U U Claim. The above definition for ω does not depend on the choice of oriented pa- rameterization φ . 1 Proof. Let φ i : U i → V, i = 1 , 2, be oriented parameterizations. Let ω ∈ Ω n c ( V 1 ∩ V 2 ). Define U 1 , 2 = φ − 1 ( V 1 ∩ V 2 ) , (6.90) 1 U 2 , 1 = φ − 1 ( V 1 ∩ V 2 ) , (6.91) 2 which are open sets in R n ....
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lecture35 - Lecture 35 Before moving on to integration we...

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