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dif9-page5 - -1 to f q x around x(p We obtain functions g...

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Proof. Since ------------------------- 1 1 d f(t x ) n i i d f f( x ) - f( 0 ) = ----------------------------------- dt = S x -------------------- (t x ) dt = j j i d x dt i=1 i 0 0 1 1 n i d f i d f = S x -------------------- (t x ) dt, we may take g ( x ) = -------------------- (t x ) dt. 44444 i j d x i j d x i=1 i i 0 0 Now we are ready to prove the theorem. Let us take a differentiable manifold (M, A ) and a chart x = (x ,. ..,x ) e A 1 n defined in a neighborhood of p e M. Define the derivations d (p) as follows i -1 d f q x # $ d (p) (f) := ------------------------------ ( x (p)). 3 i 4 d x i We prove that the derivations d (p) form a basis in the space of derivations i at p. They are linearly independent since if we have n S a d (p) = 0, i i i=1 then applying this derivation to the j-th coordinate function x we get j n d x j S a -------------------- ( x (p)) = a = 0. i d x j i=1 i On the other hand, if D is an arbitrary derivation at p, then we have n D = S D(x ) d (p). i i i=1 Indeed, let f e F (M) be an arbitrary smooth function on M and apply lemma 3
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Unformatted text preview: -1 to f q x around x (p). We obtain functions g defined around x (p) such that i-1 n d f q x f = f(p) + S (x -x (p))g q x and g ( x (p)) =------------------------------( x (p)) . i i i i d x i=1 i In this case however we have n D(f) = D(f(p)) + S D((x -x (p))) g ( x (p)) + (x (p)-x (p)) D(g q x ) = i i i i i i i=1-1 n d f q x n # $ S D(x )------------------------------( x (p)) = S D(x ) d (p) (f). i d x i 3 i 4 i=1 i i=1 To finish the proof, we only have to show that every derivation at the point p can be obtained as a speed vector of a curve passing through p. Define the curve g :[-e , e ]----------L M by the formula-1 g (t) := x ( x (p) + (t a ,...,t a )). 1 n n d Then obviously the speed vector g ’(0) is just S a--------------------(p). 44444 i d x i=1 i 5...
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