# Proof of the existence of a nontrivial Killing vector field is equivalent to the existence of a nontrivial $S^1$-action

Where can I find proof of the following classical fact?

The existence of a nontrivial Killing vector field on a compact Riemannian manifold $$M$$ is equivalent to the existence of a nontrivial $$\Bbb S^1$$-action on $$M$$.

Is there any counterexample in non-compact case?

• For the second question, consider the manifold $\mathbb{R}$ with the standard metric has a killing vector field $\frac{\partial}{\partial x}$, but $\mathbb{R}$ has no nontrivial continuous $S^{1}$-actions. To see this, note that any non-trivial orbit of an $S^{1}$-action is homeomorphic to $S^{1}$, but no subset of $\mathbb{R}$ is homeomorphic to $S^{1}$. The first question is very interesting and I hope it gets answered! – Nick L Dec 14 '18 at 15:38

## 1 Answer

The group $$\text{Isom}(M)$$ has the form of a Lie group; if $$M$$ is compact, this Lie group is compact.

Given any non-trivial Killing vector field, this gives a map $$X: \Bbb R \to \text{Isom}(M)$$; its image is a commutative subgroup. The closure of the image of $$X$$ is still commutative (the equation $$ab = ba$$ is true on an open dense subset of $$\bar X \times \bar X$$) and still a subgroup. If $$M$$ is compact, this subgroup must then be compact; because $$\bar X$$ is a compact, connected, non-trivial abelian Lie group we thus have $$\bar X \cong T^n$$ for some $$n>0$$. In particular, there is a circle subgroup of $$\text{Isom}(M)$$, and hence a faithful action of $$S^1$$ on $$M$$ by isometries.