TITLE: Infinitesimal generator of the Brownian motion on a sphere
QUESTION [5 upvotes]: As explained  here, the infinitesimal generator of a 1D Brownian motion is $\frac{1}{2}\Delta$. As discussed here, for the Brownian motion on circle we can write
$$Y_1=\cos(B) \\
Y_2= \sin(B)$$
and its Ito formula is
$$dY_1=-\frac12 Y_1 \, dt-Y_2 \, dB, \\
dY_2=-\frac12 Y_2 \, dt+Y_1 \, dB,$$
Then, to find its generator (as discussed here) we can write
$$\mathcal{A}=\frac12\left(-y_1\partial_{y_1}-y_2\partial_{y_2}+y_2^2\partial_{y_1}^2+y_1^2\partial_{y_2}^2 \right)=\frac12\partial_\theta^ 2.$$
I wonder what would be the three steps above for the Brownian motion on $S_2$ sphere?
Unlike above where we hade only ^one Brownian^, i.e., $B$, should we start with
$$Y=( \cos(B_1)\sin(B_2), \sin(B_1)\sin(B_2), \cos(B_1))$$
where $B_1$ and $B_2$ are two independent Brownians?

Are the statements above correct?
What is the intuitive explanation of the ^two Brownians^?
How to calculate the SDE and also the infinitesimal generator from $Y$?

Thanks in advance.

REPLY [2 votes]: As an alternative approach for the derivation of the advertised conclusion (in my opinion, this is a more elementary approach). You can take a look at Example 8.5.8 in Oksendal's classical textbook on SDEs, in which the author used a random time change argument to drive the SDE satisfied by a Brownian motion on the unit sphere in $\mathbb{R}^n$ (with $n \geq 3$). This SDE is written in the cartesian coordinates. However, when we consider the special case $n = 3$, a simple Ito-type computations enable us to transform the SDE in cartesian coordinate to the SDE in spherical coordinate $(\theta,\phi)$. I attached the detailed computations below, I hope you can follow and understand.