Massless limit to massive scalar in AdS space I was trying to solve massive scalar wave equation in AdS spacetime (or rather in BTZ). I noticed few funny things : 


*

*The $m\to 0$ limit to the solution is subtle. One of the two independent solutions diverges! But the other one solves the massless wave equation. Is it very generic case? The possible reason I have in mind is following. The massless scalar action enjoys more symmetry namely $\phi \to \phi+ const$  which the massive one does not. May be this enhanced global symmetry plays a role.. but I am not sure.

*The other thing is I tried to solve the $massless$ wave equation in BTZ background using mathematica. It couldn't solve it. But the massive one is solved very fast. This might be a technical part though, but I am a bit surprised.
 A: I usually work in the Poincare patch for which the line element is
$$
ds^2 = {{d {\vec x}^2 + dz^2} \over {z^2}} \ .
$$
Given translation invariance in $x$, one can then assume a solution of the form
$\phi(\vec x,z) = e^{i \vec k \cdot \vec x } f(z)$.  The wave equation for a massive scalar in $d+1$ dimensional AdS space then reduces to
$$
z^{d+1} \partial_z z^{-d+1} \partial_z f(z) - z^2 \vec k^2 f(z) - m^2 f(z) = 0 \ .
$$
The Mathematica command 
DSolve[z^(d + 1) D[z^(-d + 1) f'[z], z] - z^2 k^2 f[z] - m^2 f[z] == 
  0, f[z], z]
will produce an answer in terms of Bessel functions for general $m$ and for $m=0$.
I am not entirely sure what divergences the questioner has in mind.  The specific form for $\phi$ will of course depend on the coordinate system used.  In the Poincare patch, in a Euclidean setting and in the Lorentzian setting for space-like $\vec k$, there will be an exponential divergence at large $z$ for one of the two solutions.  In both the Lorentzian and Euclidean setting, at small $z$, there are generically two behaviors, $z^\Delta$ and $z^{d-\Delta}$ where I choose $\Delta$ to be the larger of the two solutions of the quadratic equation $\Delta(\Delta-d) = m^2$, assuming $m^2 \geq - d^2/4$ (the so-called Breitenlohner-Freedman bound).  In the massless case, these two fall-offs are $\Delta = d$ and $d-\Delta = 0$.  The fall-off $z^{d-\Delta}$ is then divergent in the sense that $\phi$ is no longer normalizable because of its behavior at $z=0$. 
