0

Let $M$ be a smooth manifold. Let $X \in \Gamma(TM)$ be a vector field on $M$, which vanishes at a finite number of points. (Every smooth manifold admits such a vector field).

Consider the subgroup $\{\phi \in \operatorname{Diff}(M)| \phi_*X=X \}$?

Is it always a finite dimensional Lie group? Or can it be infinite-dimensional?

Community
  • 1
Asaf Shachar
  • 25,967

1 Answers1

1

It may well be infinite dimensional. Take on the torus the vector field $X =\frac{\partial}{\partial x}$. For any $g(y)$ smooth and periodic function, the vector field $g(y) \frac{\partial}{\partial x}$ satisfies $[Y,X]=0$, so the difeomorphism induced by $Y$ ( $1$-parameter group in fact) invariates $X$.

orangeskid
  • 56,630
  • Tanks. Just to be sure, when you say $\frac{\partial}{\partial x}$ you refer to the angle coordinate vector field from one of the $\mathbb{S}^1$?. (I am thinking now on the torus as $\mathbb{S}^1 \times \mathbb{S}^1$). So each circle has a global non-zero vector field, and you just take one of those, considered as a vector field on the torus? – Asaf Shachar Dec 03 '15 at 14:57
  • @Asaf Shachar: Correct! And this should work for any product manifold. It would be interesting to find all the vector fields that commute ( the Lie bracket is $0$ ) with a given vector field. Say take the vector field corresponding to the rotation of the sphere around an axis. It is invariant under the rotations. Is it invariant under any other diffeomorphisms? – orangeskid Dec 03 '15 at 15:04