Is it reasonable to view the Brownian bridge as a kind of Brownian motion indexed by points on the circle?
The Brownian bridge has some strange connections with the Riemann zeta function (see Williams' article http://www.statslab.cam.ac.uk/~grg/books/hammfest/22-dw.ps, for example).
I'm looking for a heuristic explanation of why this might be the case. If one could interpret the Brownian bridge as described above, then the heuristic would be that Brownian motion is naturally associated with heat flow, which goes hand in hand with theta functions, which goes some way toward explaining the appearance of the zeta function.
I don't know where to begin reading about Brownian motion indexed by anything other than $\mathbb{R}^{+}$. The standard stochastic analysis texts don't really address the idea.
Many thanks.
Best Answer
The circle qua circle has no distinguished point. The Brownian bridge is tied down at the two endpoints, and its variance at any point $x \in [0,1]$ is $x(1-x)$, which equals 0 at the endpoints. So is there a way to get rid of the special point and make it into a process with the same variance at all points, while having those endpoints match up? It seems to me that's the question you'd have to answer.
Added a minute or so later: But you could single out another point to be the distinguished point, subtract the value of the BB at that point from its value at every point, and get another BB, not independent of the first one, with the same probability distribution.
So I think probably the answer is yes.