# addendum-lecture-18 - z 2 [email protected] 1 f(±& 1 z 1;z m [email protected] 1...

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On the second page of the handwritten notes of lecture 18 we had set up the map G ( t;z t 1 ( z; 0)) for z 2 C m 1 a . Here G : ( ";" ) ± C m 1 a ! R m : If f 2 C 1 ( R m ) then @ @t f ( G ( t;z )) = @ @t f t 1 ( z; 0))) = d t 1 ( z; 0)) ( X t 1 ( z; 0)) ) : Also @ @z j f ( G (0 ; ( z 1 ;:::;z m 1 ))) = @ @z j f ( z 1 ;:::;z m 1 ; 0) since 0 is the identity map. This implies that dG 0 ; 0 is bijective. Hence there exists b > 0 / ( b < ";b < a ) such that G ( C m b ) = W is open in R m ; contained in C m a and G : C m b ! W is a di±eomorphism. Let V 1 ( W ) ± = G 1 ² j V . If f 2 C 1 ( V ) then if
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Unformatted text preview: z 2 W @ @z 1 f (± & 1 ( z 1 ;:::;z m )) = @ @z 1 f (& & 1 ( G ( z 1 ;:::;z m ))) = @ @z 1 f (& & 1 (&( ’ z 1 (& & 1 ( z 2 ;:::;z m ; 0)) = @ @z 1 f ( ’ z 1 (& & 1 ( z 2 ;:::;z m ; 0))) = X ’ z 1 (& & 1 ( z 2 ;:::;z m ; 0)) f = X ± & 1 ( z ) f: This is a better description of the desired argument. 1...
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