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Proof

It is clear that we only have to find such coordinates at 0 in R n {\displaystyle \mathbb {R} ^{n}} . First we write X = ∑ j f j ( x ) ∂ ∂ x j {\displaystyle X=\sum _{j}f_{j}(x){\partial \over \partial x_{j}}} where x {\displaystyle x} is some coordinate system at 0 , {\displaystyle 0,} and f 1 , f 2 , … , f n {\displaystyle f_{1},f_{2},\dots ,f_{n}} are the component function of X {\displaystyle X} relative to x . {\displaystyle x.} Let f = ( f 1 , … , f n ) {\displaystyle f=(f_{1},\dots ,f_{n})} . By linear change of coordinates, we can assume f ( 0 ) = ( 1 , 0 , … , 0 ) . {\displaystyle f(0)=(1,0,\dots ,0).} Let Φ ( t , p ) {\displaystyle \Phi (t,p)} be the solution of the initial value problem x ˙ = f ( x ) , x ( 0 ) = p {\displaystyle {\dot {x}}=f(x),x(0)=p} and let

ψ ( x 1 , … , x n ) = Φ ( x 1 , ( 0 , x 2 , … , x n ) ) . {\displaystyle \psi (x_{1},\dots ,x_{n})=\Phi (x_{1},(0,x_{2},\dots ,x_{n})).}

Φ {\displaystyle \Phi } (and thus ψ {\displaystyle \psi } ) is smooth by smooth dependence on initial conditions in ordinary differential equations. It follows that

∂ ∂ x 1 ψ ( x ) = f ( ψ ( x ) ) {\displaystyle {\partial \over \partial x_{1}}\psi (x)=f(\psi (x))} ,

and, since ψ ( 0 , x 2 , … , x n ) = Φ ( 0 , ( 0 , x 2 , … , x n ) ) = ( 0 , x 2 , … , x n ) {\displaystyle \psi (0,x_{2},\dots ,x_{n})=\Phi (0,(0,x_{2},\dots ,x_{n}))=(0,x_{2},\dots ,x_{n})} , the differential d ψ {\displaystyle d\psi } is the identity at 0 {\displaystyle 0} . Thus, y = ψ − 1 ( x ) {\displaystyle y=\psi ^{-1}(x)} is a coordinate system at 0 {\displaystyle 0} . Finally, since x = ψ ( y ) {\displaystyle x=\psi (y)} , we have: ∂ x j ∂ y 1 = f j ( ψ ( y ) ) = f j ( x ) {\displaystyle {\partial x_{j} \over \partial y_{1}}=f_{j}(\psi (y))=f_{j}(x)} and so ∂ ∂ y 1 = X {\displaystyle {\partial \over \partial y_{1}}=X} as required.

• Theorem B.7 in Camille Laurent-Gengoux, Anne Pichereau, Pol Vanhaecke. Poisson Structures, Springer, 2013.