Straightening theorem for vector fields

In differential calculus, the domain-straightening theorem states that, given a vector field ${\displaystyle X}$ on a manifold, there exist local coordinates ${\displaystyle y_{1},\dots ,y_{n}}$ such that ${\displaystyle X=\partial /\partial y_{1}}$ in a neighborhood of a point where ${\displaystyle X}$ is nonzero. The theorem is also known as straightening out of a vector field.

The Frobenius theorem in differential geometry can be considered as a higher-dimensional generalization of this theorem.

Proof

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

${\displaystyle \psi (x_{1},\dots ,x_{n})=\Phi (x_{1},(0,x_{2},\dots ,x_{n})).}$ 

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

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

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

References

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