2
$\begingroup$

I would like to know if it is possible to compare two diffusion processes defined on the same manifold $\mathcal{M}$ but with respect to different metrics say $g_1$ and $g_2$.

Is there a way to apply a Girsanov theorem for the change of measure in order to express the different metric as a change in the drift?

I.e. Lets say we have a Brownian motion defined in $(\mathbb{R}^n,g_1)$ $$ dX_t = \sigma dW_t $$,

and a Brownian motion defined in $(\mathbb{R}^n,g_2)$ $$dY_t = \sigma d\tilde{W}_t$$.

Is there a way to express $Y_t$ in terms of the metric $g_1$ as a change in the drift of the process $Y_t$?

Update:

Following this paper page 86, diffusions on Riemannian manifolds can be defined through their generators. Thus we can write a Fokker-Planck equation for the evolution of the associated probability density

$$\partial_t p_t(x) = \mathcal{L} p_t(x),$$

where we assume that the expression conserves the integral of $p_t(x)$ with respect to the volume element of the manifold $d Vol_g$, i.e., $\int p_t(x) dVol_g = 1$.

For a Brownian motion on $(\mathcal{M},g)$, we use the Laplace-Beltrami operator associated with the metric $g$:

$$ \Delta_g p_t(x) = \frac{1}{\sqrt{\det g_{ij}} } \partial_i \left( \sqrt{\det g_{ij}} g^{ij} \partial_j p_t(x) \right) , $$

where $g_{ij}$ are the components of the metric in the local chart, and $g^{ij}$ are the components of the matrix inverse to $g_{ij}$.

So the different metrics $g_1$ and $g_2$ influence the way the probability density propagates in time for each process. I wonder if one could be able to define this difference in propagation in terms of a change in the drift of a diffusion process.

Improve this question
$\endgroup$
2
  • 1
    $\begingroup$ maybe also try the ideas developed in the context of Brownian motion in Ricci flow. Here "Brownian motion with respect to time-changing Riemannian metrics, applications to Ricci flow", they study Brownian motion over a whole family of metrics. $\endgroup$
    – Thomas Kojar
    Commented Nov 18, 2022 at 20:08
  • 1
    $\begingroup$ see here too for some computations relating flat and another metric: math.stackexchange.com/questions/4163105/… $\endgroup$
    – Thomas Kojar
    Commented Nov 18, 2022 at 20:13

1 Answer 1

Reset to default
2
$\begingroup$

The generator of the diffusion corresponding to a Riemannian metric (i.e., the diffusion process which the limit of the random walks such that their increments go along geodesics and the distribution of random directions comes from the volume form) is a second order differential operator whose symbol is the inverse matrix of the metric.

The drift contributes to the linear terms only and therefore does not change the symbol. Therefore, the difference between generators of Brownian motions corresponding to different metrics can not be expressed by a drift.

The situation changes if the distribution of random directions does not come from the Riemannian metric by a canonical construction but is and extra-information. If the distribution of random directions is the same for both Brownian motions, the symbols of both generators are the same and the difference between them is a drift.

Improve this answer
$\endgroup$
3
  • $\begingroup$ Thank you! I think I don't understand what do you mean with "distribution of random directions". Can you give me some reference? $\endgroup$
    – can't stop me now
    Commented Nov 18, 2022 at 15:45
  • 1
    $\begingroup$ How do you define a Brownian motion corresponding to a metric? A possible way is to define it by a formula which includes the riemannian metric as an ingredient. In this case you need to compare the outputs of these formulas for two different metrics, and see whether the difference is a drift. One can define it as the limit of the sequence of random walks. A reference is Jørgensen, E. Zeitschrift für Wahrscheinlichkeitstheorie und verwandte Gebiete 32(1–2), 1–64 (1975). A recent reference is my (open access) paper link.springer.com/article/10.100/s12220-021-00723-z . $\endgroup$
    – Vladimir S Matveev
    Commented Nov 21, 2022 at 16:20
  • 1
    $\begingroup$ The second way, i.e., the definition of the Brownian motion as the limit of a sequence of random walks requires the following data: at every point you should have a measure on the tangent space corresponding to the choice of random direction. This field of measures is sometimes called "distribution of random directions". If two Riemannian Brownian motions are constructed by the same distribution of random directions, then their generators have the same symbols so difference between generators is a drift. $\endgroup$
    – Vladimir S Matveev
    Commented Nov 21, 2022 at 16:25

You must log in to answer this question.