It's well known that $\frac{\sin{t\sqrt{-\Delta}}}{\sqrt{-\Delta}}\delta$, the fundamental solution of wave equation on the $\mathbb{R}^n$ can be expressed as the form \begin{equation} \lim_{t\to 0}Im(|x|^2-(t-i\epsilon))^{\frac{n-1}{2}}, \quad ~~~~~(1.1) \end{equation} Now, let $\Delta_g$ denote the Laplace-Beltrami operator on the standard sphere, we know that the fundamental solution of $\frac{\sin{tA}}{A}\delta$ becomes $$ \lim_{t\to 0}Im(\cos({it-\epsilon})-\cos\theta)^{\frac{n-1}{2}}, \quad ~~~~~(1.2) $$ here, we denote $A=\sqrt{-\Delta_{S^n}+\frac{(n-1)^2}{4}}$, and $\theta(x,y)$ is the geodesic distance 0n $S^n$from $x$ to $y$. Since the similarity between (1.1) and (1.2). I wonder if we can recover (1.1) from (1.2) by scaling the metric on the sphere.
It seems now that we need to consider $A_g=\sqrt{-\Delta_g+\frac{K(n-1)^2}{4}}$ on sphere with radius $R>1$, where $K=\frac{1}{R^2}$ is the curvature, my first question is what's the fundamental solution now? What's more, if we let $K\to 0$(i.e.,$R\to \infty$), is it true that the fundamental solution has an analytic continuation with respect to K so that we get (1.1) from (1.2)?
Let's assume first $n=3$, and consider the dilated sphere with radius $R$, now the fundamental solution becomes $$ \frac{\sin{tA}}{A}=\frac{\sqrt{K}\delta(\theta-t)}{\sin{\sqrt{K}t}} $$ which is an analytic function of $K$, so we can analytically continue $K$ to any real number, and in particular, if we let $K=-1$, then we get the fundamental solution on $\mathbb{H}^3$, which has constant curvature $-1$, if we let $K\to 0$, then we have now $$ \frac{\sin{t\sqrt{\Delta}}}{\sqrt{\Delta}}\delta_x(y)=\frac{\delta(r-t)}{t}, $$ where $r=|x-y|$ is the Euclidean distance, which is exactly the fundamental solution of in $\mathbb{R}^3$. And I think in higher dimension, such principles should also be valid.