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.
1 Answer
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.
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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 MaryCommented Dec 4, 2018 at 10:50