The rectification theorem states that if $y$ in the domain with the vector field $v(y)\neq 0$, then there is a change of coordinates for a small neighborhood of $y$ such that the vector field becomes a constant vector field.
Now suppose that a vector field $V$ has no singular points in a domain $U$. Can we rectify the field $V$ in the whole domain $U$?
Wikipedia explained that there is a "global constraint", where the trajectory starts out in a patch, and after visiting a series of other patches, comes back to the original one. If the next time the orbit loops around phase space in a different way, then it is impossible to rectify the vector field in the whole series of patches.
I don't understand Wikipedia's explanation. Could someone please shed some light on this ? Much appreciated.
Because there are counterexamples. Look for instance at the constant vector field $X = \frac{\partial}{\partial x}$ in the punctured plane $\mathbb{R}^2 \setminus\{(0,0)\} \cong \mathbb{C} \setminus \{0\}$ with variables $z = x + i \, y$. Then use the conformal (complex holomorphic) multiple sheeted map $w = \log(z)$ which maps multiple copies of $\mathbb{C} \setminus \{0\}$ to the whole plane $\mathbb{C}$ with variables $w = u + i \, v$,
$$ \log \, : \, \sqcup_{k \in \mathbb{Z}} \,\, \big(\mathbb{C} \setminus \{0\}\big)_k \to \mathbb{C}$$ Its inverse map is the conformal (covering) map $$z = e^{w}$$ $$e : \mathbb{C} \to \mathbb{C} \setminus \{0\}$$
Then the straight vector field $X = \frac{\partial}{\partial x}$ is mapped to the vector field $Y$ given by the differential equation $$\frac{d w}{dt} = e^{-w}$$ which written in real coordinates looks like \begin{align} \frac{d u}{dt} = e^{-u} \cos(v)\\ \frac{d v}{dt} = - e^{-u} \sin(v) \end{align} that is the vector field $$Y = e^{-u} \cos(v) \frac{\partial}{\partial u} - e^{-u} \sin(v) \frac{\partial}{\partial v}$$ This vector field cannot be globally made straight. Only the part of it, located on two of the infintely many horizontal strips in the plane. See the picture. The horizontal lines on the right are the trajectories of $X$. Notice the origin is removed! They are mapped by $w = log(z)$ onto the horizontally U-shaped curves on the left, located in the first two horizontal strips. The horizontally U-shaped curves on the left which are the trajectories of $Y$. The real line on the right is split into two trajectories, one coming into $0$ one going away from it. The former is mapped to the second horizontal line from the bottom on the left, the latter is mapped to the third .horizontal line from the bottom on the left. SO you can make the vector field $Y$ straight in one strip, but not everywhere on the plane.