8
$\begingroup$

A "geometric" rough path is a rough path such that $Sym(\mathbb{X}_{s,t})=\frac{1}{2}X_{s,t}\otimes X_{s,t}$. For example the Ito rough path is not geometric because $Sym(\mathbb{X}_{s,t})=\frac{1}{2}X_{s,t}\otimes X_{s,t}-\frac{1}{2}I(t-s)$ but the Stratonovich rough path is geometric.

Why the term "geometric" here? Is there some intuition I'm missing? When I think geometric, I think of geometric Brownian motion, (http://en.wikipedia.org/wiki/Geometric_Brownian_motion) or geometric series, something with an exponent. I don't see that here.

$\endgroup$
3
  • 2
    $\begingroup$ The definition I've seen goes like this: a smooth path has a canonical enhancement into a "smooth rough path". Then the geometric rough paths are the closure of the smooth rough paths, in the appropriate $p$-variation topology. So it could be that geometric rough paths have, in some sense, an enhancement that is geometrically reasonable, rather than arbitrary. $\endgroup$ May 27, 2015 at 3:44
  • $\begingroup$ Thanks for your response. I guess I don't understand, why is that condition geometrically reasonable? $\endgroup$
    – user69208
    May 27, 2015 at 3:46
  • 3
    $\begingroup$ I don't really know either. Another guess is the fact that the Stratonovich integral is more "geometrically reasonable" than Ito: if you have a submanifold $M \subset \mathbb{R}^n$ and a vector field $V$ which is tangent to $M$, the process $dX_t = V(X_t) \cdot \delta B_t$ (Stratonovich) will lie in $M$, but $dX_t = V(X_t) \cdot dB_t$ (Ito) will not. I am not sure but there may be some similar property of geometric rough paths. Anyway, I think one should think of "geometric" meaning more like "differential geometry" than "geometric series". $\endgroup$ May 27, 2015 at 3:53

2 Answers 2

17
$\begingroup$

Geometric rough paths have the property that if you want to solve an equation with values in a manifold, choose a coordinate chart, and write in local coordinates $$ dY^i = V_0^i(Y)\,dt + \sum_j V_j^i(Y)\,dX_j $$ for some vector fields $V_i$ (with the obvious abuse of notation that the solution actually depends on the choice of $\mathbb{X}$, not just on $X$), then the solution does not depend on the choice of chart. I believe that this is the reason why this terminology was chosen.

In other words, solutions to equations driven by "geometric" rough paths transform according to the usual chain rule, rather than some version of Itô's formula. This is also why people studying SDEs on manifolds tend to write everything in Stratonovich form, rather than in Itô form.

$\endgroup$
5
$\begingroup$

I agree with Martin's answer - but there were other additional and compelling reasons.

A rough path (as in the original papers) was a path in the tensor algebra over V with appropriate "p-variation". Without any further conditions one can solve differential equations driven by it; however, the solutions are canonically paths in in operators on functions on the manifold (think the enveloping algebra of the Lie algebra generated by the vector fields) (think randomly evolving heat kernels on the manifold) and are not evolving points on the manifold. Rough paths drive differential equations providing the vector fields come from a Lie algebra structure that is inherited from an associative structure (when truncated to order p etc.).

You can push the solution back down onto the manifold if you have a connection or other strong geometric structures on the manifold. The Ito integral you refer to is such an example. Solutions to differential equations driven by geometric rough paths stay in the integral surface or manifold defined by the original vector fields.

Another reason for the notion is that continuous bounded variation paths are p'-rough path dense in the geometric (and weakly geometric) paths. They are the closure of the classical paths. As all the basic results of integration, differential equations, .. are continuous in these metrics, and identities such as the change of variables identity therefore hold on closed sets of paths, they automatically hold for geometric rough paths if they hold for smooth paths. Paths for which one needs correction terms are more exotic and it is helpful to have language to distinguish them. Perhaps with hind sight, weakly geometric rough paths should be rough paths and rough paths should be diffusive rough paths? But it is too late now.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.