Diagonable Kernels over a Riemannian Surface

This question is motivated by this paper. There, they develop a stippling method which requires a kernel to be diagonal. Meaning a symmetric bilinear function

$K\colon \chi\times \chi\to \mathbb{R}$

from the Riemannian surface $\chi$ to the reals, and it diagonal that admits a Fourier expansion of the form

$K=\sum_l\lambda_l\psi_l(x)\overline{\psi_l(y)}$

where $\psi_l$ is a Fourier basis, eg $e^{2\pi i (n_1x_1+n_2x_2)}$. This Fourier expansion is diagonal in the sense that all the terms $\lambda_{l,k}\psi_l(x)\overline{\psi_k(y)}$ with $k\neq l$ are $0$.

In the case when $\chi$ is the Torus it is not hard to prove that if $K$ depends only on the distance, ie $K(x,y)=\varphi(\|x-y\|)$, then $K$ is diagonal.

Is it also true in the case when $\chi=\mathbb S^2$ and the Fourier basis is the spherical harmonics?

In which sense can this be generalized? Is there always a basis of $L^2(\chi)$ such that this happens?

-