Infinitesimal generator of the Brownian motion on a sphere 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.
 A: The generator of Brownian motion on $S^2$ (with the round metric) is $\frac12\Delta$, where
$\Delta$ is the Laplacian on $S^2$, in spherical coordinates
$$\Delta = \frac{1}{\sin\theta}\partial_{\theta}(\sin\theta\cdot\partial_{\theta})+\frac{1}{\sin^2\theta}\partial_{\varphi}^2.$$
Let
$$ X_t=\sin\theta_t\cos\varphi_t,\\ 
   Y_t=\sin\theta_t\sin\varphi_t,\\
   Z_t=\cos\theta_t.$$
Now you suggest that if $B_t^{(i)}$, $i=1,2$ are two independent Brownian motions, then
$$ d\theta_t = dB_t^{(1)},\\
   d\varphi_t = dB_t^{(2)},\tag1$$
defines a Brownian motion on $S^2$. We have
$$dZ_t=d\cos\theta_t=-\sin\theta_t\cdot dB_t^{(1)}-\frac12\cos\theta_t\cdot dt.$$
But since $\Delta\cos(\theta)=-2\cos\theta$, we have
$$d\cos\theta_t-\frac12\Delta\cos\theta_t=-\frac{3}{2}\cos\theta_t\cdot dt-\sin\theta_t\cdot dB_t^{(1)},$$
Thus $\cos\theta_t-\int_0^t\frac12\Delta\cos\theta_sds$ is not a local martingale. Therefore $\frac12\Delta$ is not the generator of your process (1), and therefore (1) does not define a Brownian motion on $S^2$.

There are many different ways to construct Brownian motion on the sphere. One of them works in Stratonovich form and reads
$$ d\mathbf{X}_t = \mathbf{X}_t\otimes d\mathbf{B}_t, \tag2$$
where $\otimes$ denotes a Stratonovich cross product and $\mathbf{B}_t$ is a 3d Brownian motion. In other words,
$$ dX_t = Y_t\circ dB^{(3)}_t - Z_t\circ dB^{(2)}_t,\\
   dY_t = Z_t\circ dB^{(1)}_t - X_t\circ dB^{(3)}_t, \\
   dZ_t = X_t\circ dB^{(2)}_t - Y_t\circ dB^{(1)}_t.$$
First of all we can check that by the Stratonovich chain rule
$$ d(X_t^2+Y_t^2+Z_t^2) = 2(X_t\circ dX_t+Y_t\circ dY_t+Z_t\circ dZ_t) = ... = 0,$$
hence $(X_t,Y_t,Z_t)$ is on $S^2$ for all $t\geq 0$ iff $(X_0,Y_0,Z_0)$ is on $S^2$. Then by the Stratonovich chain rule we obtain
$$ d\mathbf{X}_t = \frac{\partial \mathbf{x}}{\partial\theta}\circ d\theta_t + \frac{\partial \mathbf{x}}{\partial\varphi}\circ d\varphi_t,$$
By expanding everything and matching with the expressions above, we can solve for $d\theta_t$ and $d\varphi_t$ in terms of the Brownian motions:
\begin{align} 
d\theta_t&=\sin\varphi_t\circ dB_t^{(1)}-\cos\varphi_t\circ dB_t^{(2)},\\
   d\varphi_t&=\cot\theta_t\left(\cos\varphi_t\circ dB_t^{(1)}+\sin\varphi_t\circ dB_t^{(2)}\right)-dB_t^{(3)}.
\end{align}
We get this back into Itô form (please check), which leads to a drift term in $\theta_t$
\begin{align} 
d\theta_t&=\frac12\cot\theta_t dt+\sin\varphi_t dB_t^{(1)}-\cos\varphi_t dB_t^{(2)},\\
   d\varphi_t&=\cot\theta_t\left(\cos\varphi_t dB_t^{(1)}+\sin\varphi_t dB_t^{(2)}\right)-dB_t^{(3)}.
\end{align}
So now take a $C^2$ function $f(\theta_t,\varphi_t)$ and use Itô's lemma to check that
$$df(\theta_t,\varphi_t) = \frac12\Delta f(\theta_t,\varphi_t) dt + ... dB_t^{(1)}+ ... dB_t^{(2)}+ ... dB_t^{(3)}$$
(only the $dt$ term has to be calculated). This shows that the generator of the process (2) is indeed $\frac12\Delta$.
Addendum: By setting
\begin{align}
dB_t^{\theta}&=\sin\varphi_tdB^{(1)}_t-\cos\varphi_tdB^{(2)}_t,\\ 
dB^{\phi}_t&=\cos\theta_t(\cos\varphi_tdB^{(1)}_t+\sin\varphi_tdB^{(2)}_t)-\sin\theta_tdB^{(3)}_t
\end{align}
and checking that these are two independent Brownian motions, one can rewrite the process as
\begin{align} 
d\theta_t&=\frac12\cot\theta_t dt+dB_t^{\theta},\\
   d\varphi_t&=\frac{1}{\sin\theta_t}dB^{\phi}_t.
\end{align}
There is a third Brownian motion $dB^{N}_t=\mathbf{X}_t\cdot d\mathbf{B}_t$ that is normal to the sphere and therefore gets cancelled out.
A: Brownian motion on the unit sphere will be given by
$$X_t=(\sin(\theta_t)\cos(\phi_t),\sin(\theta_t)\sin(\phi_t),\cos(\theta_t))$$
where $(\theta_t,\phi_t)$ solves the system of SDEs given by
$$d\theta_t=dB_t^{\theta}+\frac{1}{2}\cot(\theta_t)dt, \qquad d\phi_t=\frac{1}{\sin(\theta_t)}dB_t^{\phi}.$$ Here $B^{\theta}$ and $B^{\phi}$ are independent Brownian motions.
See Why does Brownian motion have drift on Riemannian Manifolds? for the general case.
A: 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.


