I recently stumbled upon a proof while reading a paper in the field of my study. It says:
\begin{gather*} For\ any\ t >0,\ let\ B_{SE( 3)}^{( t)} :=\left[ B_{SO( 3)}^{( t)} ,\ B_{R^{3}}^{( t)}\right] ,\ then\ for\ any\ f\in C^{\infty }( SE( 3)) ,\\ f\left( B_{SE( 3)}^{( t)}\right) -f\left( B_{SE( 3)}^{( 0)}\right) -( 1/2)\int _{0}^{t} \Delta _{SE( 3)} f\left( B_{SE( 3)}^{( s)}\right) ds\ \\ is\ a\ local\ martingale\ w.r.t \ the\ filtration\ associated\ with\ \left( B_{SO( 3)}^{( t)}\right)_{t\geqslant 0} \ and\ \left( B_{R^{3}}^{( t)}\right)_{t\geqslant 0} .\\ \end{gather*}
It then uses the prop. 3.2.1 in Stochastic Analysis on Manifolds, Hsu, to conclude that the newly defined process is indeed a Brownian motion on SE(3). (The two processes in the bracket are already Brownian motions. You may refer to Appendix D of https://arxiv.org/pdf/2302.02277.pdf if you need more details of the setting.)
The result was somewhat expectable because Brownian motion is known to be determined by the Riemannian metric, and the metric on SE(3) here in this paper was given as the product metric of SO(3) and R^3. The block diagonalized form of the product metric made me think the Laplacian operator can be decomposed so as to operate on each component manifold, thereby somehow admitting a Brownian motion that runs independently on each component manifold.
The proof, however, appears more complex than I initially thought. I could decompose the Laplacian, but that alone does not seem to be a meaningful hint to construct a sequence of stopping time that satisfies the local martingale condition. In particular, I cannot see how to relate the decomposed result to the given two local martingales(due to prop. 3.2.1 of Hsu and the fact that B_SO(3) and B_R^3 are Brownian motions) below.
\begin{equation*} g\left( B_{SO( 3)}^{( t)}\right) -g\left( B_{SO( 3)}^{( 0)}\right) -( 1/2)\int _{0}^{t} \Delta _{SO( 3)} g\left( B_{SO( 3)}^{( s)}\right) ds\\ h\left( B_{R^{3}}^{( t)}\right) -h\left( B_{R^{3}}^{( 0)}\right) -( 1/2)\int _{0}^{t} \Delta _{R^{3}} h\left( B_{R^{3}}^{( s)}\right) ds\ \end{equation*}
So, I would like to ask for the proof that can fill the gap of my idea and the assertion this paper makes. Any hints or resources regarding this specific problem or even more general problem, the Brownian motion on product manifolds with the product metric, will be appreciated.
Thank you in advance.