Suppose we have a Riemannian or Lorentzian manifold with two Killing vector fields $X,Y$ such that $[X,Y]\neq 0$. Does this imply existence of a third linearly independent Killing vector field $Z$?
I know that $[X,Y]$ is itself a Killing vector. But e.g. for $\mathbb{H}^2$, one can choose the Killing vectors to satisfy $[H,J_+]=J_+$, $[H,J_-]=-J_-$ and $[J_+,J_-]=2H$, so we see that $[X,Y]$ does not generally provide a linearly independent Killing vector.
A minimal counter-example to my claim would be to find a manifold with exactly two Killing vectors $X,Y$ satisfying $[X,Y]\neq 0$, but I seem to be unable to find such a manifold.
EDIT: my original claim included further requirements on $Z$, namely that $[Z,X]\neq 0$ and $[Z,Y]\neq 0$. These follow IF we can prove the existence of a linearly independent Killing vector $Z$. Then, if e.g. $[Z,X]=0$ and $[Z,Y]\neq 0$, consider $Z^\prime = Z+Y$ which is a linearly independent Killing vector with $[Z^\prime,X]=[Y,X]\neq 0$ and $[Z^\prime,Y]=[Z,Y]\neq 0$. Finally, if both $[Z,X]=0$ and $[Z,Y]= 0$, consider $Z^{\prime\prime}=Z+X+Y$, so that $[Z^{\prime\prime},X]=[Y,X]\neq 0$ and $[Z^{\prime\prime},Y]=[X,Y]\neq 0$.
