A^{(\zeta)}, A^{(\zeta')}\right)
= - i \int d^{p+1}{\bf x} \sqrt{-g}f^{-1}g^{\mu\nu}
\left( \overline{A_\mu^{(\zeta)}}\partial_t A_\nu^{(\zeta')}
- A_\nu^{(\zeta')} \partial_t \overline{A_\mu^{(\zeta)}} \right)\,.
\label{explicit}$$

It is important to note that pure-gauge modes are orthogonal to any mode satisfying $G=0$ and, as a result, $\nabla_\nu F^{\nu \mu} = 0$. This can be shown as follows. Suppose that $A^{(\zeta)}_\mu$ and $A^{(\zeta')}_\mu$ satisfy the condition $G=0$. Then since $\Pi^{(\zeta)\mu\nu} =-\left.F^{\mu\nu}\right|_{A_\mu = A_\mu^{(\zeta)}}$, and similarly for $A_\mu^{(\zeta')}$, the inner product (\[KGinner\]) can be written as $(A^{(\zeta)}, A^{(\zeta')})=(A^{(\zeta)}, A^{(\zeta')})_{\rm inv}$ where $$(A^{(\zeta)}, A^{(\zeta')})_{\rm inv}
\equiv i \int_\Sigma d\Sigma_\mu \left[ \overline{A_\nu^{(\zeta)}}
F^{(\zeta')\nu\mu}
- \overline{F^{(\zeta)\nu\mu}}A_\nu^{(\zeta')}\right]\,.  \label{Newinner}$$

Now, let $A_\mu^{(\zeta)} = \nabla_\mu \Lambda^{(\zeta)}$ be a pure-gauge mode. Then $$(A^{(\zeta)}, A^{(\zeta')})_{\rm inv}
= i \int_\Sigma d\Sigma_\mu 
F^{(\zeta')\nu\mu}\nabla_\nu\overline{\Lambda^{(\zeta)}}
= i \int_\Sigma d\Sigma_\mu \nabla_\nu \left( \overline{\Lambda^{(\zeta)}}\;
F^{(\zeta')\nu\mu}\right) = 0  \label{invar}$$ since $F^{(\zeta')\nu\mu}$ is anti-symmetric.[^3] Thus, $(A^{(\zeta)},A^{(\zeta')}) = 0$ if $A^{(\zeta)}_\mu$ is a pure-gauge mode and $A^{(\zeta')}$ is a physical or pure-gauge mode, i.e., $$(A^{(3n;\,\omega lm)}, A^{(3n';\,\omega' l'm')}) = 0\,,
\label{A3A3}$$ $$(A^{(3n;\,\omega lm)}, A^{(1n';\,\omega' l'm')}) = 0\;, \;
(A^{(3n;\,\omega lm)}, A^{(2n';\,\omega' l'm')}) = 0\,.
\label{ApA3}$$ Next let us examine the inner product of the non-physical solutions. By letting $$R^{(0n)}_{\omega l}(r) \equiv \sqrt{f}\; r^{-p/2}\varphi^{(0n)}_{\omega l}(r)\,,$$ we find from Eq. (\[eqrA\_t\]) $$\left( \omega^2 + \frac{d^2}{{dr^*}^2} - V_0(r^*) \right)
\varphi^{(0n)}_{\omega l} 