Big Picture: What is the connection of Malliavin calculus with differential geometry? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T13:16:49Z http://mathoverflow.net/feeds/question/16504 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/16504/big-picture-what-is-the-connection-of-malliavin-calculus-with-differential-geome Big Picture: What is the connection of Malliavin calculus with differential geometry? vitp 2010-02-26T12:18:05Z 2010-03-01T09:10:31Z <p>I know that Paul Malliavin was heavily influenced by ideas from differential geometry while developing his calculus on Wiener space. But what are the concrete analogies between both areas of mathematics? What has this to do with Hörmander's theorem (if so)?</p> http://mathoverflow.net/questions/16504/big-picture-what-is-the-connection-of-malliavin-calculus-with-differential-geome/16625#16625 Answer by George Lowther for Big Picture: What is the connection of Malliavin calculus with differential geometry? George Lowther 2010-02-27T19:13:33Z 2010-02-27T19:13:33Z <p>I can't speak for Paul Malliavin's influences, but I do know a bit about Hormander's theorem (by no means, an expert), and it is naturally suited to differentiable manifolds involving largely the idea of pullbacks of vectors. Malliavin calculus was apparently initiated to give a probabilistic proof of Hormander's theorem.</p> <p>Following Rogers and Williams, Hormander's theorem concerns stochastic differential equations of the form $$\partial X = \sum_q U_q(X) \partial B^q + W(X)\partial t.$$ Here, X is a stochastic process taking values in R<sup>n</sup>, U<sub>q</sub> and W are smooth vector fields, B<sup>q</sup> are Brownian motions and &part; represents the Stratonovich integral. As Stratonovich integration satisfies the standard change of variables formula, this SDE makes sense on an arbitrary differentiable manifold. Next, according to the statement of Hormander's theorem, let [.,.] be the usual Lie Bracket for vector fields, and let A<sub>0</sub>,A<sub>1</sub>,... be the sequence of Lie algebras defined as follows. \begin{align} &amp; A_0={\rm Lie}(U_1,U_2,...),\ &amp; A_k={\rm Lie}([U,W]\colon U\in A_{k-1}) \end{align} Then Hormander's theorem states that if &cup;<sub>n</sub>A<sub>n</sub> spans the tangent space at each point of R<sup>n</sup>, then X has smooth transition densities.</p> <p>Hormander's theorem is naturally a statement concerning diffusions on differentiable manifolds, as everything I said above makes perfect sense, and is true, if R<sup>n</sup> is replaced by any differentiable manifold.</p> <p>The idea behind the Malliavin proof is to consider differentiating with respect to perturbations of the Brownian motions B<sup>q</sup>. The point is, that if B<sup>q</sup> has a small bump applied at time t, then this creates a small bump in the solution X proportional to U<sub>q</sub> at this time, which will then be propagated along the solution. In fact, solutions to the original SDE with all different points give rise to stochastically moving frames, and bumps in the solution X are transported along with these frames in a similar as vector fields give rise to transport of vectors along these fields.</p> <p>The solution for X is smooth with respect to smooth bumps in the Brownian motions, which can be shown by converting bumps in the Brownian motion into changes in the probability measure, using Girsanov transforms. So, according to Malliavin calculus, you can always differentiate the solution with respect to the Brownian motions. The idea behind the proof of Hormanders theorem, is to invert this process of varying the Brownian motion&rarr;bumps in the final position of X. To do this, it is necessary to invert the process of transporting along the moving frames. That is, it must be a 1-1 map on the tangent spaces. Then, by a stochastic "pull-back" on the manifold (or R<sup>n</sup>), you can interpret differentiating the solution with respect to its position at any time in terms of differentiating with respect to the Brownian motions.</p> <p>So, this method of proving Hormander's theorem requires you to be able to differentiate with respect to the Brownian motion (i.e., Malliavin calculus) and the rest is all differential geometry (i.e., pullbacks of tangent vectors).</p>