Straightening theorem for vector fields

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

The theorem is essentially the special case of Frobenius theorem in differential geometry.

Proof

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

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

$\Phi$ (and thus $\psi$) is smooth by smooth dependence on initial conditions in ordinary differential equations. Since

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

by uniqueness, we have ${\partial \over \partial x_{1}}\psi (x)=f(\psi (x)).$ It also follows that the differential $d\psi$ is the identity at $0$. Thus, $y=\psi ^{-1}(x)$ is a coordinate system at $0$. Finally, since $x=\psi (y)$, we have: ${\partial x_{j} \over \partial y_{j}}=f_{j}(\psi (y))=f_{j}(x)$ and so

${\partial \over \partial y_{1}}=\sum _{j}{\partial x_{j} \over \partial y_{j}}{\partial \over \partial x_{j}}=X$

as required.