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 $y_1,\dots ,y_n$ such that $X=\partial /\partial y_1$ in a neighborhood of a point where $X$ 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 ${ℝ}^n$. First we write $X=\sum _j{f}_j(x)\frac{\partial }{\partial x_j}$ where $x$ is some coordinate system at $0,$ and $f_1,f_2,\dots ,f_n$ are the component function of $X$ relative to $x.$ Let $f=(f_1,\dots ,f_n)$. By linear change of coordinates, we can assume $f(0)=(1,0,\dots ,0).$ Let $\mathrm{\Phi }(t,p)$ be the solution of the initial value problem $\dot{x}=f(x),x(0)=p$ and let

$\psi (x_1,\dots ,x_n)=\mathrm{\Phi }(x_1,(0,x_2,\dots ,x_n)).$

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

$\frac{\partial }{\partial x_1}\psi (x)=f(\psi (x))$,

and, since $\psi (0,x_2,\dots ,x_n)=\mathrm{\Phi }(0,(0,x_2,\dots ,x_n))=(0,x_2,\dots ,x_n)$, 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: $\frac{\partial x_j}{\partial y_1}=f_j(\psi (y))=f_j(x)$ and so $\frac{\partial }{\partial y_1}=X$ as required.

References

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