Let $X$ and $Y$ be smooth vector fields on $\mathbb{T}^2$.
Definition 1: Let $A$ be an open subset of $\mathbb{T}^2$, we say that $X$ and $Y$ are equivalent in $A$, if there exists a homeomorphism $h: A \to A$ such that maps orbits of $\left.X\right|_{A}$ in orbits of $\left.Y\right|_A$, preserving the orientation of the orbits.
Definition 2: We say that $X$ and $Y$ are equivalent if $X$ and $Y$ are equivalent in $\mathbb{T}^2$.
Then my question arises
My Question: If there exists $A,B \subset \mathbb{T}^2$ open subsets of $\mathbb{T}^2$ such that:
- $X$ and $Y$ are equivalent in $A$ and $B$.
- $A \cup B = \mathbb{T}^2$.
Is it true that $X$ and $Y$ are equivalent?
This seems true but I wasn't able to find a way to prove this proposition.
Can anyone help me?
This is not true. As an example, with $\mathbb{T}^2 = \mathbb{R}^2/\mathbb{Z}^2$, take $X = (1,0)$ the horizontal unit vector field, so that all orbits of $X$ are closed (circles), and take for $Y=(1,a)$ a constant vector field with an irrational slope $a$, so that $Y$ has no closed orbits at all. Obviously $X$ and $Y$ are not equivalent. However, if you choose two proper open subintervals $I,J \subset \mathbb{T}^1 = \mathbb{R}/\mathbb{Z}$ with $I \cup J = \mathbb{T}^1$, and define $A = I \times \mathbb{T}^1$ and $B = J \times \mathbb{T}^1$, then $X$ and $Y$ are easily seen to be equivalent on $A$ and $B$, since they are both equivalent to the horizontal flow on a cylinder.