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  Brownian motion over manifolds

+ 0 like - 1 dislike
174 views

Let $B(t)$ be a brownian motion over a compact riemannian manifold $(M,g)$, then have we the limit:

$$\lim_n \frac{1}{n} \sum_{k=1}^n f(B(kx))=\int_M f d\mu_g$$

for fast any $x$, and for $f$ any continuous function?

asked Feb 1 in Mathematics by Antoine Balan (-35 points) [ revision history ]
edited Feb 7 by Antoine Balan

You haven't defined the measure $\mu$, so the answer is no. Consider $M=\Bbb R$, and choose $\mu$ to be the Dirac-measure $\delta_0$. Your Wiener-process (what I suppose your Brownian motion to be) will explore the real line and sample the function $f$ thereon (assuming you meant to write $lim_{n\to\infty}$), while the integral will be $f(0)$.

Is your manifold required to be connected?

Isn't your question connected to ergodicity? 

f must be continuous. The mesure is defined because the manifold is riemannian.

I suppose you meant to write a comment instead of an answer.

$\mu$ is not defined in your question, because the presence of a metric $g$ (I assume you mean $g$ to be a metric) allows you to define a measure, but it is not clear from your question that $\mu$ has to be this measure, i.e. the canonical volume form or the corresponding Lebesgue measure. It is not forbidden to use other measures, e.g. the Dirac-measure, on manifolds with a metric.

Maybe you should put more care in your questions. Assuming, of course, that you are interested in answers.

The manifold has to be compact, otherwise take a limit of integrals.

Again you probably wanted to write a comment. I wrote my "answer" before you added the requirement of compactness. Though I don't see why this should make a difference. Maybe you aimed for something like $\mu(M)<\infty$.

After your edit of 07 or 08 Feb the problem is more specific. It reminds me of Monte Carlo integration. You should multiply the l.h.s. by $\mu_g(M)$ (consider the case of $f$ constant on $M$). $M$ should be connected, otherwise your random trajectory may not be able to leave the component of connectedness it started in. Roughly speaking the requirement is that the probability of finding your Brownian motion in a subset $U\subset M$ is proportional to $\mu_g(U)$. I would be surprised if no results on this issue were known.

From a more physical intuition, a particle diffusing in a box (connected compact manifold, the boundary shouldn't be an issue) will "forget" where it started. As all parts of the box are equivalent, the particle will visit all partial volumes of equal size with equal probability. In this case I am inclined (without having worked out a proof) to assume your equation to hold (with the modification mentioned above). 

2 Answers

+ 0 like - 0 dislike

Let $I_1:=(0,1)$, $I_2=(a,a+1)$, where $a\geq 1$, be open intervals of ${\Bbb R}$. Let $M:=I_1\cup I_2$. Choose $$f:M\rightarrow{\Bbb R},\qquad x\mapsto\cases{0 &if $x\in I_1$\cr 1&if $x\in I_2$\cr}$$

$f$ is continuous. Choose the metric to be the standard Euclidean one. Choose $\mu$ to be the one-dimensional Lebesgue-measure (i.e. ${\rm d}\mu={\rm d}x$). 

In this case, the r.h.s. of your equation will be equal to $1$. Your l.h.s. refers to a single sample trajectory of the Wiener-process. If this trajectory, considered as a trajectory in ${\Bbb R}$, starts in $I_1$, it cannot enter $I_2$. So the l.h.s. will be equal to $0$.

answered Feb 3 by Flamma (80 points) [ no revision ]
+ 0 like - 2 dislike
As the limit is a linear form over the continuous functions, it is an integral by the theorem of Riesz.
answered Feb 4 by Antoine Balan (-35 points) [ no revision ]

That is not an answer to your question. What limit are you referring to? A pointwise limit? A limit under some norm on a function space? The Riesz representation theorem requires a Hilbert space and continuous linear forms.

Note that the limit in your question is the limit of a sum the terms of which depend on the Wiener process; you are sampling function values and average over them. 

Do you wish to claim by your "answer" that because "the limit" is a linear form and there is the Riesz representation theorem, the equality in your original question is correct?

Your "answer" is not a reaction to my "answer", nor does it show any of my comments so far to be incorrect. 

The answer to your original question is "no".

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