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Jul 23, 2013 at 20:56 comment added shu @GeorgeLowther, yes, this is exact what I would like to do! But not so evident...
Jul 23, 2013 at 20:37 answer added Did timeline score: 7
Jul 23, 2013 at 20:35 comment added George Lowther This comes down to discrete approximation of a continuous diffusion. You will be guaranteed weak convergence as long as the generator of the discrete process approximates the generator of BM on the manifold (i.e., the Laplacian) closely enough.
Jul 23, 2013 at 16:33 comment added shu In fact, in the case if we take account of length of each edge, the triangulation will converge as metric space to the manifold $X$. So if we can define a random walk(which take account of length of the edge), the limit should be universal.
Jul 23, 2013 at 16:24 comment added shu What I mean by random walk is not to say at each point, we take 1/2 chance to one of the other next points. In the case you metioned, we shoud take account of length of the edge to modifier the proba to the next point.
Jul 23, 2013 at 16:21 comment added Noah Stein The limit process is not universal. Consider triangulating a circle by slicing the top half into $M$ segments and the bottom into just one. Then the scale of the motion in the top half will be $1/M$ times the scale of the motion in the bottom half at every iteration, so also in any reasonable limit.
Jul 23, 2013 at 15:53 comment added shu @Noah Stein, but the limit process might be a universal objet...
Jul 23, 2013 at 15:44 comment added Noah Stein I'm not quite sure what you want out of the construction you propose. Certainly it depends on the triangulation, whereas Brownian motion doesn't, right?
Jul 23, 2013 at 15:40 history edited shu CC BY-SA 3.0
added 362 characters in body
Jul 23, 2013 at 3:15 comment added Theo Johnson-Freyd Hi shu, Since Nate has marked this as a possible duplicate, perhaps you might expand your question to clarify why Donsker's theorem doesn't do what you want?
Jul 22, 2013 at 22:01 comment added shu hi, Nate Eldredge. I am afraid Donsker's theorem is nothing to do with the triangulation.
Jul 22, 2013 at 21:36 review Close votes
Jul 23, 2013 at 3:15
Jul 22, 2013 at 21:30 history edited Nate Eldredge
probability tag
Jul 22, 2013 at 21:20 comment added Nate Eldredge See Reference needed: Donsker's Invariance Principle for Riemannian Manifolds; I think the references given there should answer your question. I'm tentatively marking this as duplicate.
Jul 22, 2013 at 19:24 history asked shu CC BY-SA 3.0