Using the definition of the the operator to prove that the generator for Bronwnian motion on $S^2$ is $\frac{1}{2}\Delta$

Question

I am reading the book Applied Stochastic Analysis and the author claims the following:

The SDE for Brownian motion on the unit sphere $S^{n-1}$ has the Ito form $$dX_t=-\frac{n-1}{2}X_tdt+(I-X_t\otimes X_t)dW_t, X(0)=X_0\in S^{n-1}$$

where $\otimes$ is the Stratonovich product. The operator $\mathcal{L}$ of the process is $$-\frac{n-1}{2}x\cdot\nabla f(x)+\frac{1}{2}\Delta f(x)-\frac{1}{2}x\otimes x:\nabla^2 f(x)$$

where $x:\nabla^2 f(x)=\sum_{i,j}\partial_{ij}x_{ij}f$. When $n=3$, the Fokker-Planck/forward Kolmogorov equation is $$\partial_tp=\frac{1}{2}\Delta_{S^2}p$$ where $\Delta_{S^2}$ is the Laplace-Bertrami operator of the polar form $$\Delta_{S^2}=\frac{1}{\sin\theta}\frac{\partial}{\partial\theta}\left(\sin\theta\frac{\partial}{\partial\theta}\right)+\frac{1}{\sin^2\theta}\frac{\partial^2}{\partial^2\phi}, (\theta,\phi)\in[0,2\pi)\times[0,2\pi)$$

I have trouble deducing the generator for Brownian motion on the unit circle is $\frac{1}{2}\Delta$ by using the definition given in the book. Does anyone have an idea how to deduce it? I have searched other posts, like Infinitesimal generator of the Brownian motion on a sphere, but they generally used other methods. And most of the books just take the fact for granted. Thank you.

Answer 1

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 $n \geq 3$). Once the SDE is derived, the associated infinitesimal generator can be easily read from it. Let me know if you have any further questions regarding the time change argument presented in the book.