# Is the action of diffeomorphisms transitive on the set of vector fields with prescribed zero locus?

Let $M$ be a manifold, and $\sf X, \sf Y$ vecor fields with the same set of zeros: i.e., for every $x\in M$, $\mathsf{X}(x)=0$ iff $\mathsf{Y}(x)=0$. Denote $V=\{ x\in M \mid \mathsf{X}(x)=0 \}$.

Is there a diffeomorphism $\varphi:M\to M$ such that $\varphi_{*}\sf X = \sf Y$? What if $\varphi$ is required to fix $V$ pointwise?

• No this is not true. Imagine the gradient flow of the height function on the circle. And a flow with two equilibria that always points counterclockwise. – Thomas Rot Jul 8 '17 at 18:00
• Another example is the hopf fibration on The threesphere, and a flow on the threesphere without equilibria where not all orbits are periodic – Thomas Rot Jul 8 '17 at 18:02
• @ThomasRot: Oh yes, I see. Your two comments are two distinct ways in which things can go wrong. – Qfwfq Jul 8 '17 at 18:05
• The order of vanishing in an isolated zero is another additional invariant which may prevent vector fields from being related by a diffeomorphism. Think about $M=\mathbb R^n$, $V=\{0\}$ and the fields $|x|^{2k}\tfrac{\partial}{\partial x_1}$ for different $k$. – Andreas Cap Jul 9 '17 at 8:33
• @Qfwfq Aside from closed orbit or singularity, one can introduce two vector field on the plane which does not have singularity but they are not smooth related. For example : $X= \frac{\partial}{\partial x}$ and $Y=cos y \frac{\partial}{\partial x} +sin y \frac{\partial}{\partial y}$. The first vector field admits a global tranversal section but the second one does not. – Ali Taghavi Jul 10 '17 at 19:10