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Suppose we have a Brownian motion(or transition density) on a Lie group G and a Riemmanian manifold H on which G acts transitively and isometrically. Can we construct a Brownian motion( or transition density) on H? Any help will be useful.

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  • $\begingroup$ Well, you can always push forward the transition probability $\mu$ by the map $g\mapsto g\cdot x$ defined on $G$ with value on $M$, call the result $\mu_x$ and use it as the transition probability from $x$ of a random walk. It seems that your question is too imprecise to get real answers. $\endgroup$ – Benoît Kloeckner Jun 4 '13 at 10:33
  • $\begingroup$ I would like to say that Brownian motion exist always on any complete Riemmanian manifold, in particular on Homogeneous spaces. $\endgroup$ – shu Jun 4 '13 at 15:04
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This question is not as trivial as some people think.

One point is that a quotient of a Markov chain is not Markov in general; there is a certain condition that has to be satisfied. More precisely, if $\pi_x$ are the transition probabilities of a Markov chain on a space $X$, and $X\to \overline X,\; x\mapsto \overline x$ is a quotient map, then the quotient chain on $\overline X$ is Markov if and only if the images $\overline{\pi_x}$ and $\overline{\pi_y}$ of transition probabilities $\pi_x$ and $\pi_y$ coincide whenever $\overline x = \overline y$.

The other point is that in the presence of a group of symmetries it is natural to require that the transition probabilities of the quotient chain be invariant with respect to this group.

In your situation the quotient manifold (which I denote by $X$ instead of $H$ in order to avoid confusion with subgroups) is $X=G/K$, where $K$ is a compact subgroup of $G$ (because $G$ acts by isometries, so that point stabilizers are compact). The most standard example is, of course, provided by non-compact Riemannian symmetric spaces.

Assuming the original chain (or process) on $G$ is space homogeneous, the transition probabilities on $G$ are translates of a single probability measure $\mu$: $$ \pi_g = g\pi_e = g\mu \;. $$ Then the above conditions amount to bi-$K$-invariance of the measure $\mu$. Namely, this is a necessary and sufficient condition for the quotient chain on $X$ to be (i) Markov, and (ii) $G$-invariant.

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Fix $x \in M$. Since $G$ acts transitively and isometrically on $M$, there exists a left-invariant metric on $G$, that makes the map $\pi: g \in G \to gx \in M$ a Riemannian submersion. If you project the left invariant Brownian motion on $G$ by $\pi$, you get a Brownian motion on $M$ started at $x$.

In this construction, we can observe that the metric on $G$ may depend on $x$, but we can check that it does not if the distance is binvariant by the action of $G$ in the sense that $d(hx,ghx)=d(x,gx)$ for all $h,g \in G$ and $x \in M$.

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