Let $G \subset GL_n(\mathbb{R})$ be a compact Lie group. Let $\mathfrak{g}$ be its Lie algebra, with inner product given by $\operatorname{tr} (A A^t)$. Isometrically identify $\mathfrak{g} \subset M_n(\mathbb{R})$ with $\mathbb{R}^n$. Let $B(t)$ be the usual Brownian motion on $\mathfrak{g}$ under this identification. Using $ \exp : \mathfrak{g} \to G$, we can get a stochastic process, $\exp (B(t))$ on $G$, which looks reasonably like a Brownian motion starting at the identity matrix.
On the other hand, I am assured that for every Riemannian manifold, the Laplacian (somehow) produces a Brownian motion. I am intimidated by this construction (not having any background in Stochastic differential equations yet). If the general Brownian motion (for $G$ with the induced metric from the embedding) was the same as the one in my first paragraph then I would be very satisfied with that little bit of understanding.
Are they the same?
(Generally there is a geodesic exponential map for Riemannian manifolds, and I guess one could pose the same question. The metric above is bi-invariant, so my memory is that they agree. I'll stick to asking for this case of Lie groups because there is a "better chance" that it is true. However, I'm also interested in answers or comments about this interpretation.)
Edit:I guess there is no particular reason for them to be the same... lot's of things look like Brownian motion, but aren't (Brownian bridges, conditioning on being near a continuous function using Levy's forgery theorem, etc.). I guess I should just learn the accepted definition.