Unformatted text preview: f1 ) f * X . This proves equation (1). Similarly, using the deﬁnition of pushforward, we compute ( ( f * X )( h ◦ f1 ) ) f ( p ) = ( f * X ) f ( p ) ( h ◦ f1 ) = ( ( f * ) p ( X p ) ) ( h ◦ f1 ) = X p ( h ◦ f1 ◦ f ) = X p h = ( Xh )( p ) = ( Xh )( f1 ( f ( p ))) = ( ( Xh ) ◦ f1 ) f ( p ) . As before, since these function agree at any f ( p ) ∈ f M , and since f is a bijection, we conclude that ( f * X )( h ◦ f1 ) = ( Xh ) ◦ f1 . This proves equation (2). The proof of the preliminary lemma is therefore complete. 1...
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 Fall '10
 N.A
 Geometry, Trigraph, Vector Motors, Riemannian geometry

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