Given a Riemannian manifold $(M,g)$, the paths of a Brownian motion on it can be written as the following stochastic differential equation in local coordinates: $$ dX_t = \sqrt{g^{-1}} dB_t - \frac{1}{2} g^{ij}\Gamma^k_{ij} dt = \sigma(X_t)\, dB_t + \vec{b}(X_t) \,dt $$ where $B_t$ is an $n$ dimensional Wiener process and $g_{ij}\sigma^i_k\sigma^j_\ell=\delta_{k\ell}$.

My question is conceptual and geometric in nature: how can there be a drift term in this equation?

Algebraically, I understand, roughly speaking, that it arises from the extra temporal terms in Ito's formula. However, in a general relativity sort of way, one can consider $g$ to be a "warping of space" (say for $M=\mathbb{R}^n$), and we note $g$ is always symmetric. The metric does not depend on directions, but only on locations (unlike say for Finsler manifolds). In other words, acceleration in one direction due to the curvature also occurs in the opposite direction, meaning the effect of the curvature on the diffusion is also symmetric. So, geometrically, how can $\vec{b}$ exist, as it by definition favors some particular direction?

This is even weirder to me when I think about Riemannian normal coordinates (say at $p$), where $g_{ij}=\delta_{ij}$ and thus $\Gamma^k_{ij}=0$ at $p$. Thus, $\vec{b}=0$ at $p$, in that system. One can do this at every point. I suppose the drift would not disappear off of $p$, but it still seems odd to me that the presence of drift would not somehow be an invariant. Undoubtedly, I am missing something here.

I think something to do with the heat equation generating the SDE above, i.e. $\partial_t u = \Delta_g u/2$, may be useful.

Edit: It's useful to note that the term "drops out" of the equation for the Laplace-Beltrami operator in local coordinates (see the cross-posted version of this question on MathOverflow).

  • 2
    $\begingroup$ It might be instructive to try it with a particular example, say a round circle embedded in $\mathbb{R}^2$. You will find that without the drift term, the process won't stay on the circle. However, if you use a Stratonovich SDE instead of Ito, you will find the drift term is not there. In some sense Stratonovich is more intrinsic in geometric contexts. $\endgroup$ – Nate Eldredge Mar 20 '17 at 17:12
  • $\begingroup$ @NateEldredge Thanks for that example! I also must say I tend not to think of Stratonovich forms as much, evidently to my detriment. However, the circle case does not really help my intuition in an "unconstrained" case (e.g. $M=\mathbb{R}^4$), where $g$ should seemingly provide no directionality, which was my original problem. $\endgroup$ – user3658307 Mar 20 '17 at 18:31
  • 1
    $\begingroup$ On a manifold like $M = \mathbb{R}^4$ with the flat metric, the Christoffel symbols are all zero so your drift term will vanish. $\endgroup$ – Nate Eldredge Mar 20 '17 at 19:17
  • $\begingroup$ @NateEldredge Absolutely, but the flat metric does not correspond to any "curving of space", which generates a drift term, and which is what I am confused by :) (For instance, $M=\mathbb{R}^3$ with non-vanishing Ricci scalar curvature). $\endgroup$ – user3658307 Mar 20 '17 at 19:56

Some intuition can be gained by noting that this term appears even in the flat case if you choose curved coordinates. For example, on $\mathbb R^2$ you could choose a coordinate system near $p$ looking something like this:

curvilinear coordinates with shaded quadrants

Geometrically, it's clear that short-time Brownian motion (defined in some invariant way) starting at $p$ is more likely to end up in the red region. If you were to define $dX_t = \sqrt{g^{-1}} dB_t$ in these coordinates then the blue and red regions would be equally likely, since in coordinates they are just two sides of an axis. Thus we need some drift term pushing towards the right to compensate for the curvature of the coordinate system.

  • $\begingroup$ Hm, thank you for that. I suppose if the space was "intrinsically curved" (say, $g = \text{diag}([a,b,c])$ for some moving frame $F(x,y,z)$, so space is "warping" locally in directions given by $F$), the effect would be equivalent. And great image :) $\endgroup$ – user3658307 Jul 4 '17 at 0:09
  • 2
    $\begingroup$ I guess the key point is that the drift is purely an artifact of coordinates - this example shows that we can change the direction and magnitude of the drift term in the local coordinate expression just by changing the second-order behaviour of our coordinates. On intrinsically curved spaces there are no flat coordinates available, so we can't choose coordinates to make the apparent drift vanish everywhere - as you mentioned, the best we can do is geodesic normal coordinates about $p$, which makes the apparent drift vanish at $p$. $\endgroup$ – Anthony Carapetis Jul 4 '17 at 2:24

In stochastic differential geometry, as developed by Meyer and exposed by Emery in his book, the drift of a stochastic process may only be defined via an affine connection.

Recall that, in differential geometry, affine connections distinguish curves which have zero acceleration, or geodesics. There is no sense in saying that a curve is a geodesic, if one has not specified a connection, beforehand.

In stochastic differential geometry (more precisely, in the study of continuous-time stochastic processes in manifolds) affine connections distinguish stochastic processes which have zero drift, or martingales. There is no sense in saying that a process is a martingale (local martingale is more appropriate, perhaps) if one has not specified a connection, beforehand.

Meyer studies second order vectors in a manifold. A second order vector is a second order differential operator at a point (without a constant term). Then, an affine connection is defined as a linear mapping from second order vectors to first order vectors (which are just our usual vectors), which is the identity on first order vectors.

Intuitively, one can think of the stochastic differential $dX$, of a stochastic process $X$ on a manifold, as a random second order vector. Let us say that $\Gamma$ is an affine connection. The drift of $X$ with respect to $\Gamma$ is equal to the random vector $\Gamma(dX)$. In particular, if $\Gamma(dX) = 0$ then $X$ is a $\Gamma$-martingale!

This definition (you can find all computations in Meyer and Emery) shows that, indeed, a martingale in a manifold is a process with zero drift. The drift you are talking about is an artefact which appears when working in a general, perhaps not well-chosen, coordinate system.

Note that the above only uses an affine connection, which does not have to be metric. A Riemannian metric serves to distinguish a martingale which is a Brownian motion, but drift is not a ``metric property". -- Salem

  • $\begingroup$ Thank you for your answer, and I appreciate the connection to martingales! However, I am not certain that the drift is ``merely'' an artifact of coordinates; as the comments on the other answer notes, on manifolds with intrinsic curvature, it is often impossible to choose coordinates where the drift vanishes everywhere. We can of course choose them such that $\vec{b}$ vanishes at a specific point, but not everywhere at once. (It is also oddly unintuitive that Brownian motion is not a martingale in general... it sort of gets at the essence of my question...) $\endgroup$ – user3658307 Aug 31 '17 at 23:39
  • 1
    $\begingroup$ @ user3658307 : yes you are right. I had in mind vanishing at a specific point, in fact. With regard to Brownian motion, it is a martingale, with respect to the Levi-Civita connection. Being a martingale is not a coordinate-invariant property, (because of Itô's rule, roughly), therefore we cannot say that anything is a martingale, before giving an additional structure to the manifold, namely an affine connection. $\endgroup$ – Salem Sep 1 '17 at 3:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.