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?

  • 3
    $\begingroup$ 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. $\endgroup$ – Thomas Rot Jul 8 '17 at 18:00
  • 4
    $\begingroup$ Another example is the hopf fibration on The threesphere, and a flow on the threesphere without equilibria where not all orbits are periodic $\endgroup$ – Thomas Rot Jul 8 '17 at 18:02
  • $\begingroup$ @ThomasRot: Oh yes, I see. Your two comments are two distinct ways in which things can go wrong. $\endgroup$ – Qfwfq Jul 8 '17 at 18:05
  • 2
    $\begingroup$ 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 $k$. $\endgroup$ – Andreas Cap Jul 9 '17 at 8:33
  • 1
    $\begingroup$ @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. $\endgroup$ – Ali Taghavi Jul 10 '17 at 19:10

Let me summarize the conclusions in the comments. This is rarely true:

There are local obstructions around fixed points: For example the order of vanishing (Andreas Cap), or the Hopf index of an isolated equilibrium (more generally the Conley index) must be preserved by the diffeomorphism.

There are also global obstructions. A simple observation is the following: The diffeomorphism must map periodic orbits to periodic orbits. Hence the cardinality of the set of periodic orbits must be the same. The irrational flow on the two torus cannot be mapped to the rational flow (where all orbits are periodic).

Locally away from the equilibria the statement is true. There is a standard form around such points: This is called the flow box theorem. See for example Theorem 2 here.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.