Dirac bundle and Frobenius theoram

Question

We know that for any distribution W in TM, one can associate the Lagrangian subbundle. In this case how can I prove this statement?"This subbundle is dirac if and only if  W is involutive."

Answer 1

A Lagrangian Subbundle $L \subset \mathbb{T} M$ is defined as a Dirac structure if it is involutive. So, for any $a_1, a_2 \in \Gamma(L)$ $$[a_1, a_2, ] \in \Gamma(L)$$ Where $[\cdot, \cdot]$ is the Dorfman bracket. The Dorfman bracket in general is not a Lie bracket, in the sense that $(\Gamma(\mathbb{T}M), [\cdot, \cdot])$ is not a Lie algebra, because the bracket is not skew-symmetric, although one can show that it is skew-symmetric up to an exact form $$[a_1, a_2] + [a_2, a_1]=d(a_1,a_2) \tag{*}$$

So denote by $\text{Dir}(M)$ all the Dirac structures on $M$, you need to show that for $a_1, a_2 \in \Gamma(L)$ $$(a_1,a_2)=0$$ and by (*) one has that the bracket is skew-symmetric on $\Gamma(L)$, hence one can show that $[a_1, a_2] = -[a_2,a_1]$ and $(\Gamma(\mathbb{T}M), [\cdot, \cdot])$ can be then shown to be a Lie algebra.