(..this is almost a continuation of my last question (which got closed!)...) Let me first rewrite one of the main results of this paper, http://calvino.polito.it/~camporesi/JMP94.pdf in a coordinate system where the metric on $\mathbb{H}_n$ looks like, $ds^2 = \frac{1}{z^2}(dz^2 + \sum_{i=1}^{n-1} dx_i^2)$.

Then the attached paper explains that on $\mathbb{H}_n$ the spin-s fields are modelled by symmetric transverse traceless tensors of rank-s and we denote such fields as $h_{i_1 i_2\dots i_s}^{\lambda l m \sigma}$. We further want them to satisfy the following equation,

\begin{align} (-\nabla^2 )h^{\lambda lm \sigma}_{i_1 i_2\dots i_s} = (\lambda ^2 + \rho^2 + s)h^{\lambda lm \sigma}_{i_1 i_2\dots i_s} \end{align}

where $\rho = \frac{n-1}{2}$. The index $\lambda \geq 0$ (the lower bound coming from the requirement of square integrability) can be thought to correspond to the non-compact direction of $\mathbb{H}_n$ thought of in the hyperboloid model and the indices $lm\sigma$ corespond to the $S^{n-1}$ cross-section of it. Here $s \leq l \leq \infty$, $m = 0,1,2,3...,s$ for $n \geq 4$ and $m=0,1$ for $n = 3$ and $\sigma$ is a further degeneracy with multiplicity of $g(s) = \frac{ (2s+n-3)(s+n-4)! }{(n-3)!s!}$ (...for the special case of $n=3$, $g(0)=1$ and $g(s\geq 1) =2$...)

- Is my above understanding (from the paper) correct?

Further the asymptotic behaviour of these eigenfunctions in these coordinates can be derived to be such that,

\begin{align} h^{\lambda l m \sigma}_{i_1 i_2 \dots i_s } \sim \#z^{-s-\rho +i\lambda} + \#z^{s-\rho-i\lambda } \end{align}

(where $\#$ stands for the dependence on the coordinates orthogonal to the $z$ axis)

Now let us say that I move away from the eigenvalue equation above and ask as to what are the functions among the above which lie in the kernel of $(-\nabla^2 + a)$ (for some constant $a$) and which also goes as some $z^p$ near $z=0$.

Is this above question a valid question to ask? If yes then how can that be answered? (hopefully by using the asymptotics just stated above!) (...in case my last question is not well-defined then kindly frame and answer what you think would be the closest meaningful question..)