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.

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