TITLE: Brownian bridge interpreted as Brownian motion on the circle
QUESTION [14 upvotes]: 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.

REPLY [4 votes]: If you want an Ornstein-Uhlenbeck process (i.e. Brownian motion confined by a quadratic potential), there is no difficulty because this has a stationary distribution: start from that definition at time $0$, condition to be back at the same point at time $1$, and then extend by periodicity / wrap this on the circle. This gives a well-defined process indexed by the circle, and no point is special because we are always at equilibrium.
If you insist on Brownian motion, but do not want a special point on the circle, then there is a problem because there is no invariant probability measure. The right way to define things goes through white noise i.e. through the increments of Brownian motion. Start with a Brownian bridge $B$ on the interval $[0,1]$, continue it periodically / wrap it on the circle, and let $X_{s,t} = B_t-B_s$. This makes sense for any $s,t \in \mathbb R / \mathbb Z$, and it is easy to see that the distribution of $(X_{s,t})$ does not depend on the choice of the origin. So $(X_{s,t})$ is somehow "the" right object to look at.
From $(X_{s,t})$ you can recover $B_t$ if you want (because $B_t = X_{0,t}$), but then this depends on the choice of the origin. A better way is to say that you will identify two functions if they differ by a constant; then you will get the same equivalence class wherever you choose the origin, and that is as good a definition as any of Brownian motion on the circle. In particular, if what you want to do is stochastic calculus, the increments of the process are usually the relevant thing.
Another way to break the indeterminacy is to choose the path with average value $0$, which then coincides with the Fourier expansion version of Brownian motion (i.e., the Fourier series with independent Gaussian coefficients of appropriate variance).