Let $M = [1,\infty) \times S^2$. Equip $M$ with the metric $g = dr^2 + r^2 (\gamma + h)$ where $\gamma$ is a metric on $S^2$ and $h$ is a $(0,2)$ tensor on $M$ that satisfies $$h = O(1/r),\quad \partial h = O(1/r^2), \quad \partial \partial h = O(1/r^3) $$
In other words, the metric $g$ is asymptotic to $dr^2 + r^2 \gamma$.
It is clear that $g$ is an asymptotically flat metric on $M$ (or $\mathbb{R}^3 \setminus B_1$) if and only if $\gamma$ is of constant curvature 1.
Otherwise, in cartesian coordinates, the metric $g$ satisfies
$$g_{ij} = \delta_{ij} + O(1)$$
What I wish to know is if there exists any nontrivial conformal Killing vector field on $M$ that goes to $0$ as $r$ goes to $\infty$.
It is well known that if $g$ is asymptotically flat, i.e. if $\gamma$ is of constant curvature 1, then the answer is in the negative. This is because any nontrivial conformal Killing field on $M$ will asymptote to a conformal Killing field on Euclidean space, which we know grows like $r$. (Check York's paper titled "Conformal decomposition of symmetric tensors in the theory of gravitation").
But if $\gamma$ is not the round metric on $S^2$, how do I show that there is no nontrivial conformal Killing field that vanishes at infinity? (or come up with a counterexample?) What are the conformal Killing fields on a metric of the form $dr^2+r^2 \gamma$?
Any reference to a paper that uses metrics of this type is appreciated.
