 Solution

For the scalar wave separation in Kerr spacetime, continue to use $A=r^2+a^2$ and $s=\sin \theta$. First verify the determinant in the hint. Direct multiplication of the covariant $t,\varphi$ components givesHence the block determinant is $-\Delta {\sin }^2\theta$. Inverting this block givesThe other inverse components are $g^{rr}=\Delta /\Sigma$, $g^{\theta \theta }=1/\Sigma$, and $\sqrt{-g}=\Sigma s$. The covariant wave operator on a scalar consequently has the divergence formLet $m_{\varphi }$ denote the azimuthal mode number, to distinguish it from the axial vector $m$. Insert the mode $\Psi =e^{-i\omega t+i{m}_{\varphi }\varphi }R(r)\Theta (\theta )$ in the massless Klein-Gordon equation. Single-valuedness makes $m_{\varphi }$ an integer. The $t,\varphi$ derivatives giveWith $K(r)=A\omega -a{m}_{\varphi }$, division by the mode factor gives, on patches where $R\Theta \ne 0$,The radial and angular expressions must be opposite constants. Defining the separation constant as $\Lambda$, we obtain the two ordinary differential equationsThese equations also hold at zeros of a mode by continuity, without dividing there. Regular angular solutions are scalar spheroidal harmonics, with discrete $\Lambda ={\Lambda }_{\ell m_{\varphi }}(a\omega )$. For $a\omega =0$ the angular equation becomes the associated Legendre function equation, with $\Lambda =\ell (\ell +1)$ and $\ell \ge |m_{\varphi }|$, providing a useful check of the signs and normalization. The radial function here is exactly $R$ in the chosen ansatz, without an additional factor of $1/r$.