A useful Lemma you can use without proof

A useful Lemma you can use without proof - f-1 f X This...

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PMATH 465/665: Riemannian Geometry A useful Lemma you can use without proof for Assignment #4, problem 2[a]. Lemma. Let h C ( M ) and X Γ( TM ). Let f : M f M be a diffeomorphism. Then f * ( hX ) = ( h f - 1 ) f * X. (1) Note that this make sense, becaue h f - 1 C ( f M ) and f * X Γ( T f M ). It would not make sense to write hf * X , since the function and the vector field are on different manifolds. In addition, we have ( f * X )( h f - 1 ) = ( Xh ) f - 1 . (2) Again, this makes sense, because Xh C ( M ), so ( Xh ) f - 1 C ( f M ). Proof of lemma. Let p M , so f ( p ) f M , and ( f * ) p : T p M T f ( p ) f M . Using the linearity over R of ( f * ) p , we find ( f * ( hX ) ) f ( p ) = ( f * ) p ( h ( p ) X p ) = h ( p )( f * ) p ( X p ) = h ( f - 1 ( f ( p )))( f * X ) f ( p ) = ( h f - 1 )( f ( p ))( f * X ) f ( p ) = ( ( h f - 1 ) f * X ) f ( p ) . Since this is true for all p M , and since f is a bijection, it follows that f * ( hX
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Unformatted text preview: f-1 ) f * X . This proves equation (1). Similarly, using the definition of pushforward, we compute ( ( f * X )( h ◦ f-1 ) ) f ( p ) = ( f * X ) f ( p ) ( h ◦ f-1 ) = ( ( f * ) p ( X p ) ) ( h ◦ f-1 ) = X p ( h ◦ f-1 ◦ f ) = X p h = ( Xh )( p ) = ( Xh )( f-1 ( f ( p ))) = ( ( Xh ) ◦ f-1 ) 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 ◦ f-1 ) = ( Xh ) ◦ f-1 . This proves equation (2). The proof of the preliminary lemma is therefore complete. 1...
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