3
$\begingroup$

The Schouten bracket is an extension of the Lie bracket to multivector fields. Given a multivector field $\Lambda$ the vanishing of the Schouten bracket $[\Lambda,\Lambda]=0$ is referred to as a sort of "integrability" condition. Integrability of what exactly? In the case of a distribution of a collection of ordinary (1-)vector fields, of course, this is just the Frobenius theorem. What is the interpretation of this "integrability" for multivector fields? Tagging PDEs because even if I know nothing, I assume they are involved in the answer of this question.

$\endgroup$

1 Answer 1

4
$\begingroup$

One well-known example is the case of bivector fiels $\pi$. Then $[\pi, \pi] = 0$ is equivalent to say that $\{f, g\} = \pi(df, dg)$ is a Poisson bracket, i.e. satisfies the Jacobi identity. In this sense, it is really an integrability condition as one knows that a Poisson manifold admits a distribution by Hamiltonian vector fields, i.e.vector fields of the form $X_f = [f, \pi]$, which turns out to be integrable into a foliation by (symplectic) leaves. The integrability even can be seen to enter a second time as the symplectic leaves allow for Darboux charts being the local, i.e. integrated, version of the infinitesimal version of a symplectic form.

$\endgroup$
1
  • 1
    $\begingroup$ Yes, actually I had included the Poisson tag as I knew this was an important particular instance :) So in the Poisson case the vanishing of the bracket ensures the integrability of the Hamiltonian vector fields. I will not accept for now as I am interested in the "philosophy" of the vanishing Schouten bracket but if I don't get an answer I will accept yours! $\endgroup$
    – R Mary
    Commented Dec 4, 2018 at 10:50

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .