Where $Z$ is a Killing vector field (is this even necessary?) In case more assumptions are necessary, I have: $[Z,X] = [Z,Y] = g(Z,X) = g(Z,Y) = 0$ I want to prove $g([X,Y],Z) = 0$
I am trying to prove this without using coordinates. I kind of have a rough argument using the definition of the Lie derivative (I'm moving a vector that's always perpendicular to $Z$ through a curve in the orthogonal space to $Z$ so necessarily the difference must be in the orthogonal space) but I am not sure how to formalize it and whether I can simply put the definition inside the dot product. Are the assumptions that $Z$ is a Killing vector field necessary, and what about $[Z,X] = [Z,Y] = 0$?
edit: I forgot to mention, the metric is of the form $ds^2 = \Theta dz^2 + \hat{g}_{ab}dx^adx^b$ where nothing depends on $z$. I think that the precise name for this is polarized axial symmetry, or equivalently Ernst potential is $0$.
This seems to be false in general. Define the following vector fields on $\mathbb R^3$: $$ X = \frac{\partial}{\partial x} - y\frac{\partial}{\partial z}, \qquad Y = \frac{\partial}{\partial y} + x\frac{\partial}{\partial z}, \qquad Z = \frac{\partial}{\partial z}, $$ and let $g$ be the metric for which $\{X,Y,Z\}$ is an orthonormal basis. Then $Z$ is a Killing field for $g$, and $[Z,X] = [Z,Y] = g(Z,X) = g(Z,Y) = 0$; but $[X,Y]=2Z$, which is not orthogonal to $Z$.
The only issue here is whether the distribution consisting of vectors orthogonal to $Z$ is involutive. This example shows that the OP's hypotheses are not enough to guarantee that it is.