We can see here that the generator is an operator which can be determined for a stochastic process. But, in the answers and comments here we can see that the brownian motion on sphere can be constructed by the assumption that the generator is $\frac{1}{2}\Delta$.

  • Can anybody explain (for a novice in stochastic calculus), why can we "find" a generator for a Brownian motion in $R^n$, whereas, for the Brownian motion on sphere, first we assume something and then we construct the BM based on that generator? What is intuitively/physically/mathematically the difference between them?

  • What is the physics of the standard Brownian motion on a sphere? In $R^n$, i think, the formulation corresponds to the random motion of particles in a fluid. What about BM on $S^n$?


1 Answer 1


The problem simply is that $S^{n}$ is not a group or a Hilbert space. Therefore you cannot make sense of BM on $S^{n}$ via Levy-characterisation or as a Gaussian process. However, we know $S^{n}$ is a Riemannian manifold and it comes along with a natural notion of the Laplacian. Therefore, one of the easiest ways to define BM on $S^{n}$ is by postulating it to be the continuous time Markov process with generator $\frac{1}{2}\Delta$. But that's not the only way to do. Another way would e.g. be via approximation. Even without a group structure, one can make sense of random walks on $S^{n}$. Then akin to the argument in $\mathbb{R}^{n}$, which goes under Donsker's invariance principle or functional central limit theorem, one can show that after appropriate interpolation and rescaling this random walk will converge to some time continuous Markov process, whose generator is indeed again $\frac{1}{2}\Delta$.

  • $\begingroup$ Isn't $S^3$ (isomorphic to $SU(2)$) a group? $\endgroup$
    – user34971
    Aug 25, 2020 at 16:06
  • $\begingroup$ Yes, $S^{3}, S^{1}$ and $S^{0}$ happen to be Lie groups and there you may also work with Levy-characterisation. Yet, there is a caveat: you actually don't know what a Gaussian distribution is. However, one can drop the Gaussianity assumption and would still get a family of non-standard BMs plus linear drift, like OU processes. $\endgroup$
    – Tobsn
    Aug 25, 2020 at 19:42

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