Straightening theorem for vector fields

<h2 id="proof">Proof</h2>
<p>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
</p>

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})).}

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

∂
            
              ∂
              
                x
                
                  1
                
              
            
          
        
        ψ
        (
        x
        )
        =
        f
        (
        ψ
        (
        x
        )
        )
      
    
    {\displaystyle {\partial  \over \partial x_{1}}\psi (x)=f(\psi (x))}
  
,
<p>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.
</p>

<ul><li>Theorem B.7 in Camille Laurent-Gengoux, Anne Pichereau, Pol Vanhaecke. <i>Poisson Structures</i>, Springer, 2013.</li></ul>

Straightening theorem for vector fields open-in-new

Straightening theorem for vector fields