I have to solve the Klein Gordon equation for a scalar field, in global $AdS_3$ (covering space, with non periodic $\tau$) written as $$ ds^2=\frac{R^2}{\cos^2\rho}(-d\tau^2+d\rho^2+\sin^2\rho d\theta^2) $$ I Fourier transformed the scalar field as $$ \phi(\rho,\theta,\tau)=\sum_l\int d\omega Y_l(\cos\theta)e^{ik\omega\tau}\chi(\rho) $$ The equation for $\chi(\rho)$ is solved in terms of hypergeometric function, with arguments depending on $l, k, \omega$. Suppose now that I am able to come back to coordinate space, summing over Fourier modes the exact, regular, solution that I found and I impose that the non normalizable mode of this solution goes as a delta $\delta(\theta)\delta(\tau)$ near the boundary ($\rho=\pi/2$), that means that I want a localised source. (This constrains the arbitrary coefficients of the solution to be the inverse of all the stuff $k,l,\omega$-dependent behind the non normalisable mode, such that when I sum over Fourier modes I trivially obtain a delta).

From Klebanov Witten argument we know that we can write a regular solution using the bulk to boundary propagator $$ \phi(z,x)=\int dx'K(z,x-x')\phi_0(x') $$ with $\phi_0(x')$ the source. In case of a localised source $\delta(x')$ the solution become simply the propagator $\phi(z,x)=K(z,x)$.

Now, my question is, if I take my solution in global coordinates, I perform the correct coordinate transformation from global to Poincare', where the propagator $K$ is computed, and I sum over Fourier modes will I recover the propagator $K(z,x)$? The question is motivated by the fact that if this is true I believe that I can use the second approach to recover my results without use the machinery of sums. I hope I was clear.


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