Past exam of the mathematics course of the University of Cambridge / 2014 / iii / Paper 52 / 3 / b / ii / Solution

The change to ingoing Kerr coordinates adds functions of $r$ to $t$ and $\varphi$, and leaves $r,\theta$ unchanged. Differentiating at fixed $r,\theta ,\varphi$ therefore gives ${\partial }_{t}v=1$ and ${\partial }_t\chi =0$. Differentiating at fixed $r,\theta ,t$ gives ${\partial }_{\varphi }\chi =1$ and ${\partial }_{\varphi }v=0$. Hence the two Killing vector fields areThey remain the stationary and axial Killing vector fields; the coordinate change does not mix their generators with ${\partial }_r$.