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Unformatted text preview: Math 6455 Oct 10, 2006 1 Differential Geometry I Fall 2006, Georgia Tech Lecture Notes 13 Integration on Manifolds, Volume, and Partitions of Unity Suppose that we have an orientable Riemannian manifold ( M,g ) and a function f : M R . How can we define the integral of f on M ? First we answer this question locally, i.e., if ( U, ) is a chart of M (which preserves the orientation of M ), we define Z U fdv g := Z ( U ) f (  1 ( x )) q det( g ij (  1 ( x ))) dx, where g ij are the coefficients of the metric g in local coordinates ( U, ). Recall that g ij ( p ) := g ( E i ( p ) ,E j ( p )) , where E i ( p ) := d 1 ( p ) ( e i ) . Now note that if ( V, ) is any other (orientation preserving) local chart of M , and W := U V , then there are two ways to compute R W fdv g , and for these to yield the same answer we need to have Z ( W ) f (  1 ( x )) q det( g ij (  1 ( x ))) dx = Z ( W ) f (  1 ( x )) q det( g ij (  1 ( x ))) dx. (1) To check whether the above expression is valid recall that the change variables formula tells that if D R n is an open subset, f : D R is some function, and u : D D is a diffeomorphism, then Z D f ( x ) dx = Z D f ( u ( x ))det( du x ) dx. Now recall that, by the definition of manifolds,  1 : ( W ) ( W ) is a diffeo morphism. So, by the change of variables formula, the integral on the left hand side of (1) may be rewritten as Z ( W ) f (  1 ( x )) q det( g ij (  1 ( x )))det( d ( ) 1 x ) dx. 1 Last revised: November 23, 2009 1 So for equality in (1) to hold we just need to check that q det( g ij (  1 ( x ))) = q det( g ij (  1 ( x )))det( d (  1 ) x ) , for all x ( W ) or, equivalently, q det( g ij ( p )) = q det( g ij ( p ))det( d (  1 ) ( p ) ) , (2) for all p W . To see that the above equality holds, let ( a ij ) be the matrix of the linear transformation d (  1 ) and note that g ij = g ( d 1 ( e i ) ,d 1 ( e j )) = g ( d 1 d (  1 )( e i ) ,d 1 d (...
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