I'm dealing with the following problem:
"Given a set of $N$ points on the unit sphere, sample a new set of $M$ points having the same probability distribution."
I know that the simple 1D version can be solved e.g. with the inverse transform sampling method, but I couldn't find any reference on how to tackle this 3D equivalent on the unit sphere. Does anyone have good resources or hints about this?
Thank you in advance.
Continuous Uniform Distribution. The total surface of the unit sphere is $$ S_2=\int_0^{2\pi}\int_0^{\pi}\sin\theta\,d\theta\,d\phi=4\pi\,, $$ where $\theta,\phi$ are polar angles latitude and longitude (aka meridian). Unlike on earth, the convention here is that the north pole is at $\theta=0\,.$ One could now sample the random angles $\Theta,\Phi$ such that $$ \mathbb P(\Theta\le \theta,\Phi\le\phi)=\frac{1}{4\pi}\int_0^\phi\int_0^\theta\sin t\,dt\,dp=\frac{1}{4\pi}\phi\,(1-\cos\theta)\,. $$ The joint distribution of $\Theta$ and $\Phi$ is obviously the product of the marginal distributions $$ \mathbb P(\Theta\le\theta)=\frac{1}{2}(1-\cos\theta)\quad\text{ and }\quad\mathbb P(\Phi\le\phi)=\frac{\phi}{2\pi}\,. $$ Clearly, the meridian $\Phi$ has a uniform distribution on $[0,2\pi]$ and the latitude $\Theta$ can be sampled by drawing a uniform $U\in{\cal U}(0,1)$ and inverting the marginal distribution of $\Theta:$ $$ \Theta=\arccos(1-2U)\,. $$
Discrete Uniform Distribution. A finite number of $N$ points can be put relatively homogeneously onto the sphere when the differences in their longitudes $\phi_i$ and latitudes $\theta_i$ satisfy $$ \phi_i-\phi_{i-1}=\frac{2\sqrt{\pi}}{\sqrt{N}\sin\theta_i}\,,\quad\theta_i-\theta_{i-1}=\frac{2\sqrt{\pi}}{\sqrt{N}}\,,\quad i=1,...,N\,. $$ Clearly, \begin{align} &\mathbb P(\theta_{i-1}\le\Theta\le\theta_i,\phi_{i-1}\le\Phi\le\phi_i) \\[3mm] &=\mathbb P(\Theta\le\theta_i,\Phi\le\phi_i) -\mathbb P(\Theta\le\theta_i,\Phi\le\phi_{i-1})\\[3mm] &\quad-\mathbb P(\Theta\le\theta_{i-1},\Phi\le\phi_i) +\mathbb P(\Theta\le\theta_{i-1},\Phi\le \phi_{i-1})\\[3mm] &=\frac{\phi_i(1-\cos\theta_i)-\phi_{i-1}(1-\cos\theta_i)-\phi_i(1-\cos\theta_{i-1})+\phi_{i-1}(1-\cos\theta_{i-1})}{4\pi}\\ &=\frac{(\cos\theta_{i-1}-\cos\theta_i)(\phi_i-\phi_{i-1})}{4\pi}\,. \end{align} For large $N$ the difference $\theta_i-\theta_{i-1}$ small so that this probability is approximately equal to \begin{align} \frac{\sin\theta_i(\theta_i-\theta_{i-1})(\phi_i-\phi_{i-1})}{4\pi}=\frac{1}{N} \end{align} which is equal for all cells $[\phi_{i-1},\phi_i]\times[\theta_{i-1},\theta_i]\,.$