Addition property of Laplace-Beltrami eigenfunctions in symmetric spaces

[Originally posted on math stackexchage but have not received feedback in over a month, I'm hoping someone here could point me in the right direction].

Consider the eigenvalue equation for the Laplace-Beltrami operator on a manifold with metric $$ds^2=|K|^{-1}[d\chi^2+\sin_K^2\chi(d\theta^2+\sin^2\theta\,d\phi^2)]$$, where: $$\sin_K\chi=\left. \begin{cases} \sin(\chi),\, K>0\\ \sinh(\chi),\, K<0 \end{cases} \right.$$ i.e. the Helmholtz equation: $$(\Delta + k^2)\,Q_k(x) = 0$$ In the case $$K\rightarrow 0$$, we obtain the Euclidean metric, for which the (generalized) eigenfunctions are just $$Q_k(x)=e^{i \langle k, x\rangle}$$ (where $$\langle u,w\rangle$$ denotes the inner product), and have the property that $$Q_k(x+y) = Q_k(x)Q_k(y)$$.

The question: in the case $$K\neq 0$$, do the (generalized) eigenfunctions have the above property?

I am aware that in the case $$K\rightarrow 0$$, we can expand the eigenfunctions in spherical coordinates as spherical harmonics so that $$e^{i\langle k, x \rangle} = 4\pi\sum\limits_{\ell,m} i^\ell Y^m_l(\theta,\phi) j_l(k\, r)$$, and that in the case $$K\neq 0$$ only the radial equation is changed, for which the solutions are hyperspherical Bessel functions, however I can't seem to prove the additive property as there's no closed form for the eigenfunctions. Is there something elementary that I am missing here?

• Why do you expect an analogous property to hold? There are many coincidences in the $K=0$ case that are broken by $K\ne 0$. For instance, there is no reasonable "$+$" operation on the points of your symmetric space when $K\ne 0$. – Igor Khavkine Jan 3 at 11:45
• @IgorKhavkine I do not expect anything, I was just curious if such a property exists. Do you happen to have any references about properties of such symmetric spaces? – GreaterThanZero Jan 11 at 14:02